NEET 2020 - Physics Study Materials
Exam Duration: 120 Mins Total Questions : 100
If |V1+V2|= |V1- V2| and V2 is finite, then:
- (a)
V1 is parallel to V2
- (b)
V1=V2
- (c)
V1and V2 are mutually perpendicular
- (d)
|V1|=|V2|
From the question
\(|{ \overset { \rightarrow }{ v } }_{ 1 }+{ \overset { \rightarrow }{ v } }_{ 2 }|=|{ \overset { \rightarrow }{ v } }_{ 1 }-{ \overset { \rightarrow }{ v } }_{ 2 }|\)
Now, \(|{ \overset { \rightarrow }{ V } }_{ net }|=|{ \overset { \rightarrow }{ V } ' }_{ net }|\)
Hence \({ \overset { \rightarrow }{ v } }_{ 1 }\) and \({ \overset { \rightarrow }{ v } }_{2 }\) will be mutually perpendicular
Which of the following does not depend on the choice of the co-ordinate system?
- (a)
\(\overrightarrow { P } +\overrightarrow { Q } +\overrightarrow { R } \)
- (b)
(Px + Qx + Rx)\(\hat{i}\)
- (c)
\({ P }_{ x }\hat { i } +{ Q }_{ y }\hat { j } +{ R }_{ z }\hat { k } \)
- (d)
None of these
Components of vector depend on the choice of coordinate system.
If vectors \(=4\hat { i } -3\hat { j } +5\hat { k }\text{ and } \hat { i } -3\hat { j } +a\hat { k } \) are equal vectors, then the value of a is:
- (a)
5
- (b)
2
- (c)
-3
- (d)
-5
+5 \(\hat{k}\) =-a\(\hat{k}\)
a = -5
If \(\vec { A } \times \vec { B } =\vec { C } +\vec { D } \) , then select the correct alternative.
- (a)
\(\vec {B}\) is parallel to \(\vec { C } +\vec { D } \) .
- (b)
\(\vec{A}\) is perpendicular to \(\vec{C}\).
- (c)
Component of \(\vec{C}\) along \(\vec{A}\) = component of \(\vec{D}\) along \(\vec{A}\)
- (d)
Component of \(\vec{C}\) along \(\vec{A}\) = - component of \(\vec{D}\) along \(\vec{A}\)
According to definition of cross- product \(\vec { C } +\vec { D } \) is perpendicular to both A and B.
i.e.\(\vec { A } .\vec { C } +\vec { D } =0\) or \(\vec { A } .\vec { C } +\vec { A } .\vec { D } =0\)
or A (component of \(\vec{C}\) along \(\vec{A}\)) + A(component of Dalong \(\vec{A}\)) = 0 or component of \(\vec{C}\) along \(\vec{A}\) = - component of \(\vec{D}\) along \(\vec{A}\)
A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is:
- (a)
25°8'
- (b)
33°7'
- (c)
42°8'
- (d)
53°8'
\(R=\frac { { u }^{ 2 }sin2\theta }{ g } ;H=\frac { { u }^{ 2 }sin^{ 2 }\theta }{ 2g } \)
Hence, \(\frac { H }{ R } =\frac { sin^{ 2 }\theta }{ 2sin2\theta } \)
or \(\frac{1}{3}=\frac{1}{4}tan\theta\)
or tan θ=\(\frac{4}{3}\)
or θ=tan-1\(\left( \frac { 4 }{ 3 } \right) { 53 }^{ o }8'\)
Which of the following is the essential characteristic of a projectile?
- (a)
Initial velocity inclined to the horizontal
- (b)
Zero velocity at the highest point
- (c)
Constant acceleration perpendicular to the velocity
- (d)
None of the above
In a projectile motion, it is not necessary that initial velocity must be inclined to the horizontal. Further, only vertical component of velocity is zero at the highest point. Also, constant acceleration is perpendicular to velocity only at the highest point.
A hose lying on the ground shoots a stream of water upward at an angle of 60° to the horizontal with a velocity of 16 ms-1. The height at which the water strikes the wall 8 m away is:
- (a)
8.96 m
- (b)
10.96 m
- (c)
12.96 m
- (d)
6.96 m
UH = 16 cos 60° = 8 m s-1
Time taken to reach the wall = \(\frac{7.5}{7.5}\) = 1sec
Now, Uv = 16 sin 60° = 8√3 m s-1
A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3 m s -2 for 0.5 minutes. If the maximum height reached by it is 80 m, then the angle of projection is: (Takeg=10 ms-2)
- (a)
tan-1(3)
- (b)
\({ tan }^{ -1 }\left( \frac { 3 }{ 2 } \right) \)
- (c)
\({ tan }^{ -1 }\left( \frac { 4 }{ 9 } \right) \)
- (d)
\(sin^{ -1 }\left( \frac { 4 }{ 9 } \right) \)
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \)
\(80=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2\times 10 } \)
or u2sin2θ = 1600
Horizontal velocity = u cos θ = at
= 3 x 30 = 90ms-1
\(\frac { usin\theta }{ ucos\theta } =\frac { 40 }{ 90 } \)
or tanθ=\(\frac{4}{9}\) or tan-1\((\frac{4}{9})\)
Ratio of minimum kinetic energies of two projectiles of same mass is 4: 1. The ratio of the maximum height attained by them is also 4: 1. The ratio of their ranges would be:
- (a)
16: 1
- (b)
4:1
- (c)
8: 1
- (d)
2: 1
\({ K }_{ min. }=\frac { 1 }{ 2 } m(ucos\theta )^{ 2 }=\frac { 1 }{ 2 } { mu }_{ x }^{ 2 }\)(at the highest point)
i.e., \({ K }_{ min. }\alpha { u }_{ x }^{ 2 }\)
Given: \(\frac { \left( { K }_{ min. } \right) _{ 1 } }{ { \left( { K }_{ min. } \right) }_{ 2 } } =\frac { 4 }{ 1 } \)
\(\therefore \frac { { u }_{ x1 } }{ { u }_{ x2 } } =\frac { 1 }{ 2 } \)
Similarly, maximum height,
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } =\frac { { u }_{ y }^{ 2 } }{ 2g } \)
or \(H\alpha { u }_{ y }^{ 2 }\)
Given: \(\frac { { H }_{ 1 } }{ { H }_{ 2 } } =\frac { 4 }{ 1 } \)
or \(\frac { { u }_{ y1 } }{ u_{ y2 } } =\frac { 2 }{ 1 } \)
Now, \(R=\frac { { 2u }^{ 2 }sin\theta cos\theta }{ g } =\frac { { 2u }_{ x }{ u }_{ y } }{ g } \)
\(\therefore \frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { u }_{ x1 } }{ u_{ x2 } } \times \frac { { u }_{ y1 } }{ u_{ y2 } } =\left( \frac { 2 }{ 1 } \right) \left( \frac { 2 }{ 1 } \right) =\frac { 4 }{ 1 } .\)
Which of the following conversions is incorrect?
- (a)
1curie = 3.7 x 1010S-1
- (b)
1barn = 10-25m2
- (c)
1 quintal = 100 kg
- (d)
1 litre = 10-3 m3
1 barn = 100 fm2 = 10-28m2(∵ 1 fm = 10-15m)
Which of the following statements is incorrect regarding mass?
- (a)
It is a basic property of matter.
- (b)
The SI unit of mass is kg.
- (c)
The mass of an atom is expressed in u.
- (d)
It depends upon the temperature, pressure or location of the object in space.
Mass does not depend on the temperature, pressure or location of the object in space.
A physical quantity X is related to four measurable quantities a, b, c and d as given, X = a2b3c5/2d-2. The percentage error in the measurement of a, b, c and d are 1%, 2%, 2% and 4% respectively. What is the percentage error in quantity X?
- (a)
15%
- (b)
17%
- (c)
21%
- (d)
23%
A new system of units is proposed in which unit of mass is α kg, unit of length is β m and unit of time is γ s. What will be value of 5 J in this new system?
- (a)
5αβ2γ-2
- (b)
5α-1β-2γ2
- (c)
5α-2β-1γ-2
- (d)
5α-1β2γ2
Lightning was discovered by
- (a)
Ohm
- (b)
Thomson
- (c)
Franklin
- (d)
Faraday
Lighting was discovered by Franklin.
Which of the following is the smallest unit?
- (a)
millimetre
- (b)
angstrom
- (c)
fermi
- (d)
metre
1 mm = 10-3 m; 1 \(\mathring { A } \) = 10-10m; 1 fm = 10-15m Among the given units fermi is the smallest unit.
The ratio of the volume of the atom to the volume of the nucleus is of the order of
- (a)
1010
- (b)
1015
- (c)
1020
- (d)
1025
Radius of the atom, Ra = 1 \(\mathring { A } \) = 10-10m
Volume of the atom \({ V }_{ a }=\frac { 4 }{ 3 } { \pi R }_{ a }^{ 3 }=\frac { 4 }{ 3 } \times \pi \times { \left( { 10 }^{ -10 } \right) }^{ 3 }{ m }^{ 3 }\)
Radius of the nucleus, Rn = 1 fermi = 10-15m
Volume of the nucleus, Vn =\(\frac { 4 }{ 3 } { \pi R }_{ n }^{ 3 }=\frac { 4 }{ 3 } \times \pi \times { \left( { 10 }^{ -15 } \right) }^{ 3 }{ m }^{ 3 }\)
Their corresponding ratio is
\(\frac { { V }_{ a } }{ { V }_{ n } } =\frac { { \left( { 10 }^{ -10 } \right) }^{ 3 } }{ { \left( { 10 }^{ -15 } \right) }^{ 3 } } \frac { { 10 }^{ -30 } }{ { 10 }^{ -45 } } ={ 10 }^{ 15 }\)
Assertion: When we change the unit of measurement of a quantity, its numerical value changes.
Reason: Smaller the unit of measurement smaller is its numerical value.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
If u1 and u2 be the units to measure a quantity Q and n1 and n2 be the numerical values respectively then we know that Q = n1u1 = n2u2' Since the quantity Q does not change irrespective of the units used to measure it Q = constant. So nu = constant \(\Rightarrow constant\ n\propto \frac { 1 }{ u } i.e\)i.e. smaller the unit of measurement, greater is the corresponding numerical value
The dimensional formula of Young's modulus is:
- (a)
[ML2T-2]
- (b)
[ML-1T-2]
- (c)
[MLT-3]
- (d)
none of these
The number of particles crossing the unit area perpendicular to the x-axis per unit time is given by: \(N=-D\left( \frac { { n }_{ 2 }-{ n }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right) ,\) where n1 and n2 are the numbers of particles per unit volume for the values of x meant to be x1and x2 respectively. What is the dimensional formula for the diffusion constant D?
- (a)
[M0LT2]
- (b)
[M0L2T4]
- (c)
[M0LT-3]
- (d)
[M0L2T-1]
\(Here,\left[ n \right] =\left[ M{ ^{ 0 }L }^{ -2 }{ T }^{ -1 } \right]
\)
\( \left[ { n }_{ 1 } \right] =\left[ { n }_{ 2 } \right] =\left[ M{ ^{ 0 }L }^{ -3 }{ T }^{ 0 } \right]
\)
\(and\ \left[ { x }_{ 1 } \right] =\left[ { x }_{ 2 } \right] =\left[ M{ ^{ 0 }L }{ T }^{ 0 } \right]
\)
\(Hence,\left[ D \right] =\frac { \left[ n \right] }{ \left[ { n }_{ 1 } \right] } \times \left[ { x }_{ 1 } \right]
\)
\( =\frac { \left[ M{ ^{ 0 }L }^{ -2 }{ T }^{ -1 } \right] }{ \left[ M{ ^{ 0 }L }^{ -3 }{ T }^{ 0 } \right] } \times \left[ M{ ^{ 0 }L }{ T }^{ 0 } \right] =\left[ M{ ^{ 0 }L }^{ -2 }{ T }^{ -1 } \right] .\)
Given that \(\upsilon \) is the speed, r is radius and g is acceleration due to gravity. Which of the following is dimensionless?
- (a)
\(\upsilon \)2r/g
- (b)
\(\upsilon \)2/rg
- (c)
\(\upsilon \)2g/r
- (d)
\(\upsilon \)2rg
If C, R, Land 1 denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are:\((i)CR\quad (ii)L/R\quad (iii)\sqrt { LC } \quad (iv)L{ I }^{ 2 }\)
- (a)
(i) and (ii)
- (b)
(i) and (iii)
- (c)
(i) and (iv)
- (d)
(i), (ii) and (iii)
Capacitance, C = [M-1L-2 T 4I2]
Resistance, R = [ML2 T -3I-2]
Inductance, L = [ML2 T - 2I2]
Electric current, I = [I]
Dimensional formulae of CR, L/R and \(\sqrt { LC } \) is same as that of time.
The values of two resistors are R1= (6 ± 0.3)\(k\Omega \) and R2 = (10 ± 0.2)\(k\Omega \) . The percentage error in the equivalent resistance when they are connected in parallel is :
- (a)
5.125%
- (b)
2%
- (c)
3.125%
- (d)
10.125%
R1= (6 ± 0.3)\(k\Omega \) and R2 = (10 ± 0.2)\(k\Omega \) , Rp=?
\({ R }_{ parallel }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } \quad \left[ Let\quad ({ R }_{ 1 }+{ R }_{ 2 })=X \right] \text{ or }\quad { R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ X } \)
\(\text{In } { R }_{ p }=\ln { { R }_{ 1 } } +\ln { { R }_{ 2 } } -\ln { { X } } \)
\(\text{Differentiating, } \frac { \Delta { R }_{ p } }{ { R }_{ p } } =\frac { \Delta { R }_{ 1 } }{ { R }_{ 1 } } +\frac { \Delta { R }_{ 2 } }{ { R }_{ 2 } } +\left( -\frac { \Delta X }{ X } \right) \)
In addition or subtraction, errors are calculated as follows:
\(\Delta X \text { is mean } (\Delta { R }_{ 1 }+\Delta { R }_{ 2 })={ \Delta X }_{ mean }=\frac { 0.3+0.2 }{ 2 } =0.25\Omega \\ { R }_{ mean }=\frac { 6\Omega +10\Omega }{ 2 } =8\Omega \\ \therefore { R }_{ mean }=\frac { 6\Omega +10\Omega }{ 2 } =8\Omega \\ \frac { \Delta X }{ X } =\frac { 0.25 }{ 8 } =0.03125\ \)
\( \therefore \text{ Total error } =\frac { 0.3 }{ 6 } +\frac { 0.2 }{ 10 } +\frac { 0.25 }{ 8 }\)
= 0.05 + 0.02 + 0.03125 = 0.10125
\(\therefore \quad \frac { \Delta { R }_{ p } }{ { R }_{ p } } \approx \)10%
Inertia is that property of a body by virtue of which the body is
- (a)
unable to change by itself the state of rest
- (b)
unable to change by itself the state of uniform motion
- (c)
- (d)
unable to change by itself the state of rest or of uniform motion
Inertia means resistance to change. It is the property of the body by virtue of which it cannot change by itself its state of rest or of uniform motion.
Which of the following statements is incorrect?
- (a)
A cricketer moves his hands backwards while holding a catch
- (b)
A person falling from a certain height receives more injuries when he falls on a cemented floor than when he falls on a heap of sand.
- (c)
It is easier to push a lawn mower than to pull it.
- (d)
Mountain roads are generally made winding upwards rather than going straight up.
It is easier to pull lawn mower than to push it. All other statements are true.
A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 kg/s and at a speed of 5 mls. The initial acceleration of the block will be:
- (a)
2.5 m/s2
- (b)
5 m/s2
- (c)
10m/s2
- (d)
20m/s2
The water jet striking the block at the rate of I kg/s at a speed of 5m1swill exert a force on the block
\(F=v\frac { dm }{ dt } \) = 5 x 1 = 5N
Under the action of this force of 5 N, the block of mass 2 kg will move with an acceleration given by:
\(a=\frac { F }{ m } =\frac { 5 }{ 2 } \) = 2.5 ms-2
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U(x) = K |x|3, where K is a positive constant. If the amplitude of oscillation is a, then its time period T is:
- (a)
proportional to 1/ \(\sqrt { a } \)
- (b)
independent of a
- (c)
proportional to \(\sqrt { a } \)
- (d)
proportional to a3/2
U(x) = K(x)3
F = \(F=\frac { du }{ dx } =-3{ kx }^{ 2 }\)
Now, F = ma = m\(\frac { du }{ dt } =m\frac { dv }{ dx } \times \frac { dx }{ dt } \)
= \(mv\frac { dv }{ dx } \)
\(\therefore\) \(v\frac { dv }{ dx } =-\frac { 3K }{ m } { x }^{ 2 }\)
Integrating on both sides, we get;
\(\frac { { v }^{ 2 } }{ 2 } =-\frac { k }{ m } { x }^{ 3 }+c\)
And as for x = a, V = 0, \(\therefore\) \(c=\frac { k }{ m } { a }^{ 3 }\)
hence v = \(\frac { dx }{ dt } =\left[ \frac { 2k }{ m } \left( { a }^{ 3 }-{ x }^{ 3 } \right) \right] \)
i.e., \(dt=\sqrt { \frac { m }{ 2K } \frac { dx }{ \sqrt { { a }^{ 3 }-{ x }^{ 3 } } } } \)
So = 4t = \(4\sqrt { \frac { m }{ 2K } } \int { \begin{matrix} a \\ 0 \end{matrix} } \frac { dx }{ \sqrt { { a }^{ 3 }-{ x }^{ 3 } } } \)
put x3 = a3 sin2 \(\theta\)
\(\therefore\) dx = a \(\left( \frac { 2 }{ 3 } \right) \) sin-1/3 \(\theta\) cos \(\theta\) d \(\theta\)
\(\therefore\) T = \(4\sqrt { \frac { m }{ 2K } } \times \frac { 2 }{ 3 } -\frac { a }{ { a }^{ 3/2 } } \int { \begin{matrix} \pi /2 \\ 0 \end{matrix}sin^{ -1/3 } } \theta d\theta \)
i.e., \(T\infty \frac { 1 }{ \sqrt { a } } \)
The limiting friction is:
- (a)
always greater than the dynamic friction
- (b)
always less than the dynamic friction
- (c)
equal to the dynamic friction
- (d)
sometimes greater and sometimes less than the dynamic friction
A machine gun fires n bullets per second and the mass of each bullet is m: If v is the speed of each bullet, then the force exerted on the machine gun is:
- (a)
mng
- (b)
mnv
- (c)
mnvg
- (d)
(mnv) /g
Change in momentum in one see, i.e.,
F = change in momentum per bullet x no. of bullets fired per second
= mv x n = mnv
A block of mass 1 kg is placed on a wedge shown in figure. Find out minimum coefficient of friction between wedge and block to stop the block on it.
- (a)
0.6
- (b)
0.9
- (c)
1
- (d)
0.2
For minimum value of \(\mu\), we can write
\(m g \sin \left(45^{\circ}\right)=\left(f_s\right)_{\max }\)
\(m g \sin 45^{\circ}=(\mu)_{\min } m g \cos 45^{\circ}\)
\((\mu)_{\min }=1\)
A force of 250 N is required to lift a 75 kg mass through a pulley system. In order to.lift the mass through 3 m, the rope has to be pulled through 12 m. The efficiency of the system is:
- (a)
50%
- (b)
75%
- (c)
33%
- (d)
90%
The frictional force between two surfaces is independent of:
- (a)
nature of surface
- (b)
size of the body
- (c)
area of contact
- (d)
mass of the body
The frictional force between two surfaces,
fR = \(\mu _{ S }R={ \mu }_{ s }mg\)
where \({ \mu }_{ s }\) depends on the nature of surfaces and their materials. Thus, frictional force is independent of area of contact of the surfaces.
A piece of ice slides down a 45° incline in twice the time it takes to slide down a frictionless 45° incline. What is the coefficient of friction between the ice and incline?
- (a)
0.25
- (b)
0.50
- (c)
0.75
- (d)
0.40
In both the cases, the initial speed u and the distance s travelled is the same.
According to formula s = \(\mu\)t + \(\frac { 1 }{ 2 } \) at2
at2 is constant (as u = 0 in both the cases)
Let a1 be the acceleration of the ice block on rough incline and t1 be the time taken to slide down thisrough plane.Similary let a2 and t2 be the similar quantities in case of frictionless inclined plane.
Gievn that; t1 = 2t2
Further \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { t }_{ 2 }^{ 2 } }{ { 4t }_{ 1 }^{ 2} } \)
or \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { t }_{ 2 }^{ 2 } }{ { 4t }_{ 2 }^{ 2 } } =\frac { 1 }{ 4 } \) ........(i)
But a1 = g (sin \(\theta\) -\(\mu\)cos \(\theta\))......(ii)
and a2 = g sin \(\theta\) ................(iii)
\(\therefore\) \(\frac { g\left( sin\theta -\mu cos\theta \right) }{ gsin\theta } \) = \(\frac { 1 }{ 4 } \)
or 4 sin \(\theta\) - 4 \(\mu\) cos \(\theta\) = sin \(\theta\)
3 sin \(\theta\) = 4\(\mu\) cos \(\theta\)
\(\therefore\) \(\mu\) = \(\frac { 3 }{ 4 } \)tan \(\theta\) = \(\frac { 3 }{ 4 } \)tan450 or \(\mu\) = \(\frac { 3 }{ 4 } \) = 0.75
For ordinary terrestrial experiments the observer in an inertial frame in the following cases, is:
- (a)
a child revolving in a giant wheel
- (b)
a driver in a sports car moving with a constant high speed of 200 km/hr on a straight road
- (c)
the pilot of an aeroplane which is taking-off
- (d)
a cyclist negotiating a sharp curve
An inertial frame is a non-accelerated frame. (a), (c) and (d) have all circular motion or motion along a curved path. They have all acceleration, although in curvilinear motion, centripetal acceleration is made ineffective by the centrifugal force.
A body is moving with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is 10 m/s 2, find the minimum distance it can be stopped.
- (a)
400 m
- (b)
40 m
- (c)
0.40 m
- (d)
4 m
\(\mu\) = 72 km/h = 20 m/s, v = 0
a = \(\mu\)g = 0.5 x 10 m/s2
From, v2 = u2 -2as
\(\therefore\) (0)2 = (20)2 - 2 x 0.5 x 10 x s
\(\therefore\) x = \(\frac { 20\times 20 }{ 2\times 0.5\times 10 } \) = 40m
Assertion (A) : Newton's second law indicates that when a net force acts on an object, it must accelerate.
Reason (R) : When two or more forces are applied to an object, it must accelerate.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
- (e)
If assertion is false but reason is true.
Assertion (A) : Static limiting frictional force is always greater than the kinetic frictional force.
Reason (R) :\(\mu\)s (coefficient of static friction) > \(\mu\)k (coefficient of kinetic friction).
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
- (e)
If assertion is false but reason is true
A car is negotiating a curve of radius 150 m with a speed of 15 ms -1 . The angle through which the pendulum suspended from the top of the ceiling would deviate is: (g=10 ms-2)
- (a)
tan -1 (3/20)
- (b)
tan -1 (5/16)
- (c)
tan -1 (4/15)
- (d)
tan -1 (3/16)
The maximum velocity at which a truck can safely travel without toppling over, on a curve of radius 250 m (the height of the centre of gravity of the truck above the ground is 1.5 m and the distance: between the wheels is 1.5 m, the track being horizontal) is:
- (a)
30 ms-1
- (b)
35 ms-1
- (c)
40 ms-1
- (d)
45 ms-1
A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If its equation of motion is y = bx2 (b is a constant), its velocity component in the x-direction is :
- (a)
\(\sqrt { \frac { 2b }{ a } } \)
- (b)
\(\sqrt { \frac { a }{ 2b } } \)
- (c)
\(\sqrt { \frac { a }{ b } } \)
- (d)
\(\sqrt { \frac { b }{ a } } \)
- (e)
\(\sqrt{ba}\)
Differentiating W.r.t.t on both sides, we get,
\(\frac { dy }{ dt } =b2x\frac { dx }{ dt } \)
vy=2bxvx
Again, differentiating w.r.t. t on both sides, we get,
\(\frac { d{ \upsilon }_{ y } }{ dt } =2b{ \upsilon }_{ x }\frac { dx }{ dt } +2bx\frac { d{ \upsilon }_{ x } }{ dt } \)
=2bvx2+0
[\(\frac { d{ \upsilon }_{ x } }{ dt } =0\)because the particle has constant acceleration along y-direction]
As per question
\(\frac { d{ \upsilon }_{ y } }{ dt } =a=2b{ \upsilon }_{ x }^{ 2 };{ \upsilon }_{ x }^{ 2 }=\frac { a }{ 2b } \)
\({ \upsilon }_{ x }=\sqrt { \frac { a }{ 2b } } \)
(A) As the frictional force increases, the safe velocity limit for taking a turn on an unbanked road also increases.
(R) Banking of roads will increase the value of limiting velocity.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
- (e)
If assertion is false but reason is true.
Both assertion and reason are true but reason is not the correct explanation of assertion.
On an unbanked road, friction provides the necessary centripetal force, \(\frac { mv^{ 2 } }{ r } \)=F=μmg
∴ v=\(\sqrt { \mu rg } \)
Thus, with increase in friction, safe velocity also increases. When the road is banked with angle of θ, then its limiting velocity is given by,
\(v=\sqrt { \frac { rg(tan\theta +\mu ) }{ 1-\mu tan\theta } } \)
Thus, limiting velocity increases with banking of road.
(A) If the speed of a body is constant, the body cannot have a path other than a circular or straight line path.
(R) It is not possible for a body to have a constant speed in an accelerated motion.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
- (e)
If assertion is false but reason is true.
Angular momentum of the particle rotating with a central force is constant due to:
- (a)
constant torque
- (b)
constant force
- (c)
constant linear momentum
- (d)
zero torque
Central forces pass through axis of rotation so torque is zero. If no external torque is acting on a particle, the angular momentum of a particle is constant.
A person slides freely down a frictionless inclined plane while his bag falls down vertically from the same height. The final speeds of the man (vM) and the bag (vB) should be such that:
- (a)
uM < uB
- (b)
uM = uB
- (c)
they depend on the masses
- (d)
uM > uB
In the question number 77, the force acting on the particle is
- (a)
mω2\(\vec { r } \)
- (b)
-mω2\(\vec { r } \)
- (c)
2mω2\(\vec { r } \)
- (d)
-2mω2\(\vec { r } \)
Given: \(\vec { r } \) =A cosωt \(\hat { i } \) + B sin ωt \(\hat { j } \)
Velocity, \(\vec { v } \) =\(\frac { d\vec { r } }{ dt } =\frac { d }{ dt } \) (A cosωt \(\hat { i } \) + B sin ωt \(\hat { j } \))
= -Aω sinωt \(\hat { i } \) + Bω sin ωt \(\hat { j } \)
Acceleration, \(\vec { a } =\frac { d\vec { v } }{ dt } \) = Aω2 cosωt \(\hat { i } \) + Bω2 sin ωt \(\hat { j } \)
\(\vec { a } \) = -ω2 [A cosωt \(\hat { i } \) + B sin ωt \(\hat { j } \)] =-ω2\(\vec { r } \)
The force acting on the particle is
\(\vec { F } =m\vec { a } =m(-{ \omega }^{ 2 }\vec { r } )=-m\omega ^{ 2 }\vec { r } \)
Assertion: The familiar equation rng = R for a body on a table is true only if the body is in equilibrium.
Reason: The equality of rng and R has no connection with the third law.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
The two forces mg and R can be different. For example a body in an accelerated lift. The equality of mg and R has no connection with the third law because two forces mg and R can be different.
A mass m moves in a circle on a smooth horizontal plane with velocity V0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R0/2. The final value of the kinetic energy is
- (a)
2mv02
- (b)
1/2mv02
- (c)
mv02
- (d)
1/4mv02
Using the conservation of angular momentum, we see that:
mv0R0 = mv2 \(\times\) R0/2
or, v2=2v0
Now, KE = 1/2 \(\times\) m(2v0)2 = 2mv02
In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of 2 m S-1at 45° from a height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m S-2, the kinetic energy of the shotput when it just reaches the ground will be :
- (a)
2.5 J
- (b)
5 J
- (c)
525 J
- (d)
640 J
Using law of energy conservation,
(TME)i = (TME)f \(\Rightarrow\) Ui + Ki = Uf + Kf
\(\Rightarrow\) 20 x 10 x 3 +\(\frac{1}{2}\) x 20 x 22 = 0 + Kf \(\Rightarrow\) Kf = 640 J
An open watertight railway wagon of mass 5 x 103 kg is moving with an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change will occur in the kinetic energy of the wagon, when it has collected 103 kg of water?
- (a)
1200 J
- (b)
300 J
- (c)
600 J
- (d)
900 J
Initial momentum of the wagon = 5 x 103 x 1.2
Let v 'be the final velocity of the wagon.
Then final momentum of the wagon
= (5 x 103 + 103)v'
According to law of conservation of momentum,
(5 x 103 + 103) v' = 5 x 103 x 1.2
\(\therefore\) v' = 1 m/sec
Change in KE = initial KE - final KE
= \(\frac{1}{2}\)(5 x 103)(12)12 - \(\frac{1}{2}\) x (6 x 103)(1)2
= 3600 - 3000 = 600 J.
Assertion: Astronauts in a satellite moving around the earth are in a weightless condition.
Reason: The satellite and its contents are falling freely at the same rate.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion
- (c)
If assertion is true but reason is false
- (d)
If both assertion and reason are false
The escape velocity on a planet, four times the radius of the earth and having 9 times acceleration due to gravity is:
- (a)
67.2 km/sec
- (b)
37.4 km/sec
- (c)
403.2 km/sec
- (d)
422.2 km/sec
A satellite orbits around the earth in a circular orbit with a speed v and orbital radius r. If it loses some energy, then v and r changes as:
- (a)
- (b)
- (c)
- (d)
The condition for a uniform spherical mass mofradius r to be a black hole is: (G = gravitational constant and g = acceleration due to gravity)
- (a)
(2Gm/r)1/2 \(\le \)c
- (b)
(2gm/r)1/2 =c
- (c)
(2Gm/ r)1/2\(\ge \)c
- (d)
(gm/ r)1/2 \(\ge \) c
The criterion for a star to be black hole is:
\(\frac { GM }{ { c }^{ 2 }r } \ge \frac { 1 }{ 2 } \)
or \(\sqrt { \frac { 2Gm }{ r } } \ge c.\)
At what height h above the earth, the value of g becomes g/2? (R = Radius of the earth)
- (a)
- (b)
\(\sqrt { 2 } \)R
- (c)
(\(\sqrt { 2 } \)-1)R
- (d)
\(\frac { 1 }{ \sqrt { 2 } } \)R
The value of acceleration due to gravity at height h (when h is not negligible as compared to R )
\(g'=g\frac { { R }^{ 2 } }{ (R+h)^{ 2 } } \)
Here, \(g'=\frac { g }{ 2 } \)
\(\therefore \frac { g }{ 2 } =g\frac { { R }^{ 2 } }{ (R+h)^{ 2 } } \)
or \(\frac { 1 }{ 2 } =\frac { { R }^{ 2 } }{ (R+h)^{ 2 } } \)
or \(\sqrt { \frac { 1 }{ 2 } } =\frac { R }{ R+h } \)
or \(R+h=\sqrt { 2 } R\)
\(\therefore h=(\sqrt { 2 } -1)R.\)
(A) Moon travellers tie heavy weight at their back before landing on moon.
(R) The value of 'g' is small at moon.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false.
- (e)
The Young's modulus of brass and steel are respectively 1.0 x 1011 Nm-2 and 2.0 x 1011 Nm-2. A brass wire and a steel wire of the same length are extended by 1 mm each under the same force. If radii of brass and steel wires are RB and Rs respectively, then:
- (a)
\({ R }_{ S }=\sqrt { 2R_{ B } } \)
- (b)
\({ R }_{ S }={ R }_{ B }/\sqrt { 2 } \)
- (c)
Rs = 4RB
- (d)
Rs = RB/2
Now Young's modulus is given as
Y = Fl/AΔl
So YA = constant
Now Y1r12 =Y2r22
1.0 x RB2 = 2 x Rs2
Hence, \({ R }_{ s }={ R }_{ B }/\sqrt { 2 } \)
In case of a hollow body, if ρB and ρs represent the densities of body and substance respectively, then:
- (a)
- (b)
- (c)
- (d)
For a hollow body, as Vbody > Vsub., hence density of body is less than that of substance.
The fraction of a floating object of volume Vo and density do above the surface of a liquid of density d will be:
- (a)
\({d_o\over d}\)
- (b)
\({dd_o\over d+d_o}\)
- (c)
\({d-d_o\over d}\)
- (d)
\({dd_o\over d-d_o}\)
A ball whose density is 0.4 x 103 kg/m3 falls into water from a height of 9 cm. To what depth does the ball sink?
- (a)
9 cm
- (b)
6 cm
- (c)
4.5 cm
- (d)
2.25 cm
Velocity of ball just before entering the surface of water =\(\sqrt{2gh}=\sqrt{2\times 980\times9}cm/sec\)
This velocity is retarded by the upthrust acting on the ball due to water. The retardation
\(={V\rho g -V\sigma g \over V\rho}=({\rho - \sigma \over \rho})g=({0.4-1\over 0.4})g=-{3\over 2}g\)
Hence, depth of penetration of ball into water is given by:
02 - 2 x 980 x 9 = 2\((-{3\over2})\times 980 \times h\)
\(\therefore\) h = 6 cm.
Two equal drops are falling through air with a steady velocity of 5 cm/sec. If the drops coalesce, the new terminal velocity will be:
- (a)
5 x 2 cm/sec
- (b)
5 x \(\sqrt{2}\) cm/sec
- (c)
5 x (4)1/3 cm/sec
- (d)
\(\frac { 5 }{ \sqrt { 2 } } \) cm/sec
Construction of submarines is based on:
- (a)
- (b)
- (c)
- (d)
Newton's laws
An areoplane of mass 3 x 104 kg and total wing area of 120 m2 is in a level flight at some height. The difference in pressure between the upper and lower surface of its wings, (in kilo pascals) is:
- (a)
2.5
- (b)
5.0
- (c)
10.0
- (d)
12.5
Upward thrust = \(\frac { Fprce }{ Area } =\frac { 3\times { 10 }^{ 4 }\times 10 }{ 120 } \)=2.5 kPa.
The fraction of a floating object of volume Vo and density \(\rho\)o above the surface of liquid of density \(\rho\) will be :
- (a)
\(\rho_o\over \rho -\rho_o\)
- (b)
\(\rho -\rho_o\over \rho\)
- (c)
\(\rho_o\over \rho\)
- (d)
\(\rho_o\rho\over \rho+\rho_o\)
Let f be the fraction of volume of the object floating above the surface of liquid.
According to Archimedes' principle,
Weight of liquid displaced = Weight of object
(Vo -fVo)\(\rho\)g = Vo\(\rho\)og
or (1- f)\(\rho\) = \(\rho\)o or f = 1-\(\rho_o\over \rho\)
or \(f={\rho-\rho_o\over \rho}\)
The gate of a canal is 8 m wide. The level of water on one side is 30 m and on the other side is 15 m. The resultant force on the gate is :
- (a)
270 x 105 N
- (b)
270 x 106 N
- (c)
540 x 105 N
- (d)
540 x 106 N
\(F={1\over 2}\rho g h_1A_1-{1\over 2}\rho g h_2 A_2\)
\(={1\over2}\)x 104[ 30 x 30 x 8 - 15 x 15 x 8]
= 270 x 105 N
(A) 1 kg of cotton fibre will weight equals in air when made more fluffy.
(R) Weight of air in cotton will cancel out with the force of extra buoyancy acting on it.
- (a)
- (b)
- (c)
- (d)
- (e)
For a gas. the difference between two specific heats is 5000 J/mole-°C. If the ratio of specific heats is 1.6, the two specific heats are: (in J/mole-°C)
- (a)
- (b)
- (c)
- (d)
If one mole of a monoatomic gas \(\gamma\) =5/3 is mixed with one mole ofa diatomic gas \(\gamma\) =7/5, what is the value ofy for the mixture?
- (a)
1.5
- (b)
1.53
- (c)
1.60
- (d)
1.52
\(\gamma\)=\(\left( \frac { { C }_{ p } }{ { C }_{ v } } \right) _{ average }\)
For MA gas: Cv=\(\frac{3R}{2}\)
For DA gas: Cv=\(\frac{5R}{2}\)
∴ (Cv)av=\(\frac { \frac { 3R }{ 2 } +\frac { 5R }{ 2 } }{ 2 } \)=2R
Also, (Cp)av=(Cv)av+R=3R
\(\gamma\)=\(\left( \frac { { C }_{ p } }{ { C }_{ v } } \right) _{ average }=\frac { 3R }{ 2R } \)=1.5
For an ideal monoatomic gas, the universal gas constant R is n times the molar heat capacity at constant pressure Cp. Here n is:
- (a)
0.67
- (b)
1.4
- (c)
0.4
- (d)
1.67
For ideal monoatomic gas,
Cp=\(\frac{5R}{2}\)
or R=\(\frac{2}{5}\)Cp=0.4 Cp
∴ n=0.4
An ideal heat engine works between temperatures T1=500K and T2 = 375K. If the engine absorbs 600 J of heat from the source, then the amount of heat released to the sink is:
- (a)
- (b)
- (c)
- (d)
\(\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
or Q2=\(\frac { { T }_{ 2 }{ Q }_{ 1 } }{ { T }_{ 1 } } =\frac { 375\times 600 }{ 500 } \)=450 J
N molecules, each of mass m, of gas A and 2 N molecules, each of mass 2 m, of gas B are contained in the same vessel which maintained at a temperature T. the mean square of the velocity of molecules of B type is denoted by v2 and the mean square of the X component of the velocity of A type is denoted by ω2, then (ω2/ v2) is:
- (a)
2
- (b)
1
- (c)
1/3
- (d)
2/3
Mean square velocity of gas molecules is
v2= 3kT/m
Now for gas B:
\({ v }_{ B }^{ 2 }\) = v2 = 3kT/2m
Since the molecule has equal probability of motion in all directions, so:
\({ v }_{ x }^{ 2 }={ v }_{ y }^{ 2 }={ v }_{ z }^{ 2 }={ \omega }^{ 2 }\)
Now, \({ v }^{ 2 }={ v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 }+{ v }_{ z }^{ 2 }=3{ v }_{ x }^{ 2 }\)
\({ v }_{ x }^{ 2 }=\frac { { v }^{ 2 } }{ 3 } \)
ω2 = 1/3 x 3kT/m = (kT/m)
\(\frac { { \omega }^{ 2 } }{ { v }^{ 2 } } \) = [(kT/m) / (3kT/2m)] = 2/3
The equation of a simple harmonic wave is given by Y = 5 sin \(\frac{\pi}{2}\) (l00 t - x), where x and yare in metre and time is in second. The time period of the wave (in seconds) will be
- (a)
0.04
- (b)
0.01
- (c)
1
- (d)
5
Y = 5 sin \(\frac{\pi}{2}\) (l00 t - x )
y = A sin (\(\omega\)t - kx)
\(\therefore \omega \)\(\frac { 2\pi }{ \lambda } =\frac { 2\pi }{ 0.08 } =25\pi \)
\(\beta =\omega =\frac { 2\pi }{ T } =\frac { 2\pi }{ 2 } =\pi \)
Which of the following statements is incorrect for a stationary wave?
- (a)
Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
- (b)
All the particles cross their mean position at the same time.
- (c)
All the particles are oscillating with same amplitude.
- (d)
There is no net transfer of energy across any plane.
Two pendulums begin to swing simultaneously. The first pendulum makes 9 full oscillations when the other makes 7. The ratio of lengths of the two pendulums is:
- (a)
9/7
- (b)
7/9
- (c)
49/81
- (d)
81/49
\(\cfrac { { t }_{ 0 } }{ 9 } =2\pi \sqrt { \cfrac { h }{ g } } \) and \(\cfrac { { t }_{ 0 } }{ 7 } =2\pi \sqrt { \cfrac { h }{ g } } \)
\(\therefore \) \(\cfrac { h }{ { l }_{ 2 } } =\left( \cfrac { 9 }{ 7 } \right) ^{ 2 }=\cfrac { 49 }{ 81 } \)
Two sound waves having a phase difference of 60° have a path difference of:
- (a)
2\(\lambda \)
- (b)
\(\lambda \)/2
- (c)
\(\lambda \)/6
- (d)
\(\lambda \)/3
An echo repeats 2 syllables. If the speed of sound is 330 m s -1, then the distance of the reflecting surface is:
- (a)
- (b)
- (c)
- (d)
A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is:
- (a)
qEy2
- (b)
qE2y
- (c)
qEy
- (d)
q2Ey
Kinetic energy = Force x Displacement
So, as per the question, kinetic energy attained by
the particle after moving a distance y will be qEy
Two identical capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other:
- (a)
The charges on the free plates connected together are destroyed
- (b)
The charges on the free plates are enhanced
- (c)
The energy stored in the system increases
- (d)
The potential difference in the free plates becomes 2V
Now Q1 = CV
Q2 = CV
As per conservation of charge
CV1 + CV2 = Q1 + Q2
Also CV1 + CV2 = 2CV
Hence V1 + V2 = 2V
A copper cylindrical tube has inner radius a and outer radius b. The resistivity is \(\rho \) . The resistance of the cylinder between the two ends is
- (a)
\(\frac { \rho l }{ { b }^{ 2 }-{ a }^{ 2 } } \)
- (b)
\(\frac { \rho l }{ 2\pi (b-a) } \)
- (c)
\(\frac { \rho l }{ \pi ({ b }^{ 2 }-{ a }^{ 2 }) } \)
- (d)
\(\frac { \pi \left( { b }^{ 2 }-{ a }^{ 2 } \right) }{ \rho l } \)
If one had considered a solid cylinder of radius b, one can suppose that it is made of two concentric cylinders of radius a and the outer part, joined along the length concentrically one inside the other.
If Ia and Ix are the currents flowing through the inner and outer cylinders
\(\because { I }_{ total }={ I }_{ b }={ I }_{ a }+{ I }_{ x }\Rightarrow \frac { V }{ { R }_{ b } } =\frac { V }{ { R }_{ a } } +\frac { V }{ { R }_{ x } } \)
where Rb is the total resistance and Rx is the resistance of the tubular part.
\(\therefore \quad \frac { 1 }{ { R }_{ x } } =\frac { 1 }{ { R }_{ b } } -\frac { 1 }{ { R }_{ a } } \)
But \({ R }_{ a }=\frac { \rho l }{ \pi { a }^{ 2 } } \) and \({ R }_{ b }=\frac { \rho l }{ \pi { b }^{ 2 } } \)
\(\therefore \quad \frac { 1 }{ { R }_{ x } } =\frac { \pi }{ \rho l } ({ b }^{ 2 }-{ a }^{ 2 })\) \(\therefore \quad { R }_{ x }=\frac { \rho l }{ \pi ({ b }^{ 2 }-{ a }^{ 2 }) } \)
Assertion: A girl sits on a rolling chair, when she stretch her arms horizontally, her speed is reduced.
Reason : Principle of conservation of angular momentum is applicable in this situation.
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion
- (c)
If assertion is true but reason is false
- (d)
If both assertion and reason are false.
If friction in the rotational mechanism is neglected, there is no external torque about the axis of rotation of the chair and hence lω is constant, where 1 is moment of inertia and ω is angular velocity. Stretching the arms increases 1 about the rotation, results in decreasing the angular speed ω. Bringing the arms closer, body has the opposite effect.
In the following question, a statement of assertion is followed by a statement of reason. Mark the correct choice as :
Assertion: Drift velocity of electrons is independent of time.
Reason : Electrons are accelerated in the presence of electric field
- (a)
If both assertion and reason are true and reason is the correct explanation of assertion.
- (b)
If both assertion and reason are true but reason is not the correct explanation of assertion
- (c)
If assertion is true but reason is false.
- (d)
If both assertion and reason are false
Drift velocity is the average velocity of electrons in presence of electric field, which is independent of time.
A circular loop of radius R carrying a current I is placed in a uniform magnetic field B perpendicular to the loop. The force on the loop is
- (a)
2\(\pi\)RIB
- (b)
2\(\pi\)RI2B3
- (c)
\(\pi\)R2IB
- (d)
zero
If a long straight wire carries a current of 40 A, then the magnitude of the field B at a point 15 ern away from the wire is
- (a)
5.34 x 10-5 T
- (b)
8.34 x 10-5 T
- (c)
9.6 x 10-5 T
- (d)
10.2 x 10-5 T
1= 40 A
r = 15 cm = 15 x 10-2 m
\(\therefore B=\frac { { \mu }_{ 0 }I }{ 2\pi R } =\frac { 2\times { 10 }^{ -7 }\times 40 }{ 15\times { 10 }^{ -2 } } =\frac { 80 }{ 15 } \times { 10 }^{ -5 }\)=5.34x10-5 T
The vertical component of earth's magnetic field at a place is \(\sqrt { 3 } \) times the horizontal component the value of angle of dip at this place is
- (a)
30°
- (b)
45°
- (c)
60°
- (d)
90°
\(As,\quad { B }_{ V }=\sqrt { 3 } { B }_{ H }\\ Also\ \tan { \delta = } \frac { { B }_{ V } }{ { B }_{ H } } =\frac { \sqrt { 3 } { B }_{ H } }{ { B }_{ H } } =\sqrt { 3 } \quad 0r\ \delta =\tan { ^{ -1 } } \quad (\sqrt { 3 } )={ 60 }^{ 0 }\\ \therefore \ Angle\quad of\ tip,\ \delta ={ 60 }^{ 0 }\)
A jet plane is travelling west at the speed of 1600 km h-1. The voltage difference developed between the ends of the wing having a span of 20 m, if the earth's magnetic field at the location has a magnitude of 5 x 10-4 T and the dip angle is 30° is :
- (a)
4.1 V
- (b)
2.2 V
- (c)
3.2 V
- (d)
3.8 V
Here, v = 1600 km h-1 = 1600 x \(\frac{5}{18}\)= 444 m S-1
= 20 m , B = 5 x 10-4 T and \(\delta \) = 30°
As \(\varepsilon \) = Blv = Vlv
(where Vis the vertical component of earth's field)
= (B sin \(\delta \)) lv (\(\therefore\)V = Bsin8)
= 5 x 10-4 sin 30° x 20 x 444 = 2.2 V
In a pure capacitive circuit if the frequency of ac source is doubled, then its capacitive reactance will be
- (a)
remains same
- (b)
doubled
- (c)
halved
- (d)
zero
Capacitive reactance,
\({ X }_{ c }=\frac { 1 }{ 2\pi \upsilon c } \)
\(\Rightarrow\) \({ X }_{ c }\propto \frac { 1 }{ \chi } \) \(\therefore\) \(\frac { { X }_{ { C }_{ 1 } } }{ { X }_{ { C }_{ 2 } } } \)= \(\frac { { \upsilon }_{ 2 } }{ { \upsilon }_{ 1 } } \) = \(\frac { 2\upsilon }{ \upsilon } \) = 2
\(\Rightarrow\) \({ X }_{ { C }_{ 2 } }=\frac { { X }_{ { C }_{ 1 } } }{ 2 } \)
In a series LCR circuit, the phase difference between the voltage and the current is 45°. Then the power factor will be
- (a)
0.607
- (b)
0.707
- (c)
0.808
- (d)
1
Here, \(\phi \)= 45°
In series LCR circuit, power factor = cos\(\phi \)
\(\therefore cos\phi =cos45^{ 0 }=\frac { 1 }{ \sqrt { 2 } } =0.707\)
An electromagnetic wave travelling along z-axis is given as E = Eo cos(kz-ωt).Choose the correct options from the following:
- (a)
The associated magnetic field is given as \(\vec { B } =\frac { 1 }{ 2 } \hat { k } \times \vec { E } =\frac { 1 }{ \omega } (\vec { k } \times \vec { E } )\)
- (b)
The electromagnetic field can be written in terms of the associated magnetic field as \(\vec { E } =c(\vec { B } \times \hat { j } )\)
- (c)
\(\hat { k } .\vec { E } =0.\hat { k } ,\vec { B } \neq 0\)
- (d)
\(\hat { k } .\vec { E } =0,\hat { k } \times \vec { B } =0\)
We know that, direction of propagation of electromagnetic wave is always along the direction of
vector product \((\vec{E} \times \vec{B})\)
\(\vec { B } =B\hat { j } =B(\hat { k } \times \hat { i } )=\frac { E }{ c } (\hat { k } \times \hat { i } )\Rightarrow \vec { B } =\frac { 1 }{ c } [\hat { k } \times E\hat { i } ]=\frac { 1 }{ c } [\hat { k } \times \vec { E } ]\)
If we write the equation in terms of magnetic field, then
\(\vec{E}=c(\vec{B}\times \vec{k})\)
\(\hat { k } .\vec { E } =\hat { k } .(E\hat { i } )=0,\hat { k } .\hat { B } =\hat { k } .(B\hat { j } )=0\)
as angle between them is 90°.
Two slits are separated by a distance of 0.5 mm and illuminated with light of l.=6000A. If the screen is placed 2.5 m from the slits. The distance of the third bright image from the centre will be :
- (a)
1.5 mm
- (b)
3 mm
- (c)
6 mm
- (d)
9 mm
Let the distance of 11th bright fringe from centre, yn/ = nλD/d
Distance of 3rd bright image from centre,
y3 = 3 x 6000 x 10-10 x 2.5/0.5 x 10-3
= 9 x 10-3 m = 9 mm
Assertion: The stopping potential depends on the frequency of incident light.
Reason : The stopping potential is related to maximum kinetic energy by e V0 = Kmax.
- (a)
- (b)
- (c)
- (d)
The stopping potential is related to the maximum kinetic energy of electrons emitted as
Rutherford's experiments suggested that the size of the nucleus is about
- (a)
10-14 m to 10-12 m
- (b)
10-15 m to 10-13 m
- (c)
10-15 m to 10-14 m
- (d)
10-15 m to 10-12 m
How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as \(^2_1 H +^2_1H \rightarrow ^3_2He + n + 3.27 \)MeV
- (a)
2.4 x 106 years
- (b)
7.4 x 104 years
- (c)
1.6 x 106 years
- (d)
4.9 x 104 years
Number of atoms present in 2g of deuterium=6.023 x 1023
Total number of atoms present in 2000g of deuterium
=\({6.023 \times 10^{23}\times 2000 \over 2}=6.023 \times 10^{26}\)
Energy released in the fusion of 2 deuterium atoms =3.27 MeV
Total energy released in the fusion of 2.0Kg of deuterium atoms
\(E={3.27\over2}\times 6.023 \times 10^{26}=9.81 \times 10^{26}MeV\)
=15.696 x 1013J
Energy consumed by the bulb per second =100J
Time for which the bulb will glow
t=\({15.69 \times 10^{13}\over 100}s \ or \ t={15.69 \times 10^{11}\over 3.15 \times 10^7}\) years
=4.9 x 104 years.
Assertion: According to electromagnetic theory an accelerated particle continuously emits radiation.
Reason: According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic.
- (a)
- (b)
- (c)
- (d)
According to classical electromagnetic theory, an accelerated charge continuously emits radiation. As electrons revolving in circular paths are constantly experiencing centripetal acceleration, hence they will be losing their energy continuously and the orbital radius will go on decreasing and form spiral and finally the electron will fall on the nucleus.
Choose the only false statement from the following:
- (a)
In conductors, the valence and conduction bands may overlap
- (b)
Substances with energy gap of the order of 10 eV are insulators
- (c)
The resistivity of semiconductor increases with increase in temperature.
- (d)
The conductivity of semiconductor increases with increase in temperature
Option (a) is correct as in conductor conduction and valance band overlap and the conduction band is partially filled.
Option (b) is correct as insulators have energy gap of 5 -10 eV.
Option (c) is incorrect as resistivity decreases with increase in temperature.
Option (d) is correct as with increase in temperature, more and more electrons jump to conduction band, hence conductivity increases.
When n-type semiconductor is heated:
- (a)
number of electrons increases while that of holes decreases
- (b)
number of holes increases while that of electrons decreases
- (c)
number of electrons and holes remain same
- (d)
number of electrons and holes increases equally
Due to heating, when a free electron is produced then Simultaneously a hole is also produced.
The direction of the angular velocity vector is along
- (a)
the tangent to the circular path
- (b)
the inward radius
- (c)
the outward radius
- (d)
the axis of rotation
The direction of the angular velocity is along the axis of rotation.
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