IISER Physics - Description of Motion in Two and Three Dimension
Exam Duration: 45 Mins Total Questions : 30
A,B,C and D are points in a vertical line such that AB = BC = CD.If a body falls from rest at A, then the times of decent through AB,BC and CD are in the ratio
- (a)
\(1 : (\sqrt2 - 1) : (\sqrt3 - \sqrt2)\)
- (b)
\((\sqrt2 - 1) : 1 : (\sqrt3 - \sqrt2) \)
- (c)
\((\sqrt2 - 1) : (\sqrt3 - \sqrt2) : 1\)
- (d)
None of these
Let h - Ab = BC = CD. Let t1, t2, t3 be the time taken in covering distances AB, BC and CD respectively.
For distance AB, h = \(\frac{1}{2}\) \({ t }_{ 1 }^{ 2 }\)
For AC, 2h = \(\frac{1}{2}\) g ( t1 + t2 )2
For AD, 3h = \(\frac{1}{2}\) g ( t1, t2, t3 )2
Hence t1 : t2 + t1 : t1 + t2 + t3 = 1 : \(\sqrt { 2 } \) : \(\sqrt { 3 } \)
t1 : t2 + t3 = t1 : (t2 + t1) - t1 : (t1 + t2 + t3) - ( t2 + t1)
= 1 : \(\sqrt { 2 } \) - 1 : \(\sqrt { 3 } \) - \(\sqrt { 2 } \)
A ball is rolled off along the edge of table (horizontal) with velocity 4 ms-1.It hits the ground after time 0.4 s.Which one of the following is wrong
A: The height of the table is 0.8 m
B: It hits the ground at angle of 60° with the vertical
C: It covers a horizontal distance 1.6 m
D: It hits the ground with vertical velocity 4 ms-1
- (a)
If A and C are correct
- (b)
If B and D are correct
- (c)
If A and B are correct
- (d)
If C and D are correct
Let θ be the angle at which the ball hits the ground with the vertical
tan θ = \(\\ \frac { { v }_{ x } }{ { v }_{ v } } =\frac { 4 }{ gt } =\frac { 4 }{ 10\times 0.4 } =1\)
or θ = 45o
Velocity on striking the ground is
v = \(\sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 } } \) = \(\sqrt { { u }^{ 2 }+{ u }^{ 2 } } =\sqrt { 2u } \)
= 4\(\sqrt { 2 } \) ms-1
If retardation produced by air resistance is g/10, then time of flight of projectile will nearly
- (a)
Increases by 1%
- (b)
decreases by 1%
- (c)
remain same
- (d)
decreases by 2%
When no retardation, then \(T_0={2u\ sin\ \alpha\over g}\)
When air retardation of \(g\over 10\) exist, then
\(T={u\ sin\ \alpha\over g+{g\over 10}}+{u\ sin\ \alpha\over g-{g\over 10}}\)
\(T={2u\ sin\ \alpha\over g}.({{5\over 11}+{5\over 9}})\)
\(T=T_0\times5\times {20\over 99}={100\over 99}T_0\)
So, percentage increases = \(({100\over 99}-1)\times 100\)
Percentage increases = 1%
A shot is fired from a point at a distance of 200 m from the foot of a tower 100 m high so that it just passes over it. The direction of shot is:
- (a)
30°
- (b)
45°
- (c)
60°
- (d)
70°
Hmax. =100m,Rmax. =2x200=400m
\(∴\ tan\theta={4 \times100\over 400}=1\)
ө=45°
If the time of flight of a projectile is doubled, what happens to the maximum height attained?
- (a)
Halved
- (b)
Remains unchanged
- (c)
Doubled
- (d)
Becomes four times
From above question \({h\over T^2}={g\over 8}\)
If T is doubled, h becomes four times
A particle is kept at rest at the top of a sphere of diameter 42 m. When disturbed slightly, it slides down. At what height h from the bottom, the particle will leave the sphere?
- (a)
14m
- (b)
28m
- (c)
35m
- (d)
7m
The range R of projectile is same when its maximum heights are h1 and h2. What is the relation between R, h1 and h2?
- (a)
\(R=\sqrt{h_1h_2}\)
- (b)
\(R=\sqrt{2h_1h_2}\)
- (c)
\(R=2\sqrt{h_1h_2}\)
- (d)
\(R=4\sqrt{h_1h_2}\)
Range is same for angles of projection θ and 90° - θ
\(∴\ R={u^2sin2\theta\over g};h_1={u^2sin^2\theta\over g}\ and\ h_2={u^2cos^2\theta\over 2g}\)
Hence, \(\sqrt{h_1h_2}={u^2sin\theta cos\theta\over 2g}\)
\(={1\over 4}\left[u^2sin\theta\over g\right]={R\over 4}\)
Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is π/3 and its maximum height is hi then the maximum height of the other will be
- (a)
3h1
- (b)
2h1
- (c)
h1/2
- (d)
h1/3
\({h_2\over h_1}={u^2sin^2\theta_2\over 2g}\times {2g\over u^2sin^2\theta1}\)
\(={sin^1\theta_2\over sin^2\theta_1}={sin^2\pi/6\over sin^2\pi/3}={1\over 3}\)
∴ h2=h1/3
A projectile is projected with initial velocity \((6\hat i+8\hat j)\) rn/sec. If g = 10ms-2, then horizontal range is:
- (a)
4.8 metre
- (b)
9.6 metre
- (c)
19.2 metre
- (d)
14.0 metre
Here, u cos θ = 6, u sin θ = 8 and θ = 10
\(R={u^2sin2θ\over g}={2u^2sinθ cosθ\over g}\)
\(={2(usinθ)(ucosθ)\over g }={2\times8\times6\over 10}=9.6m\)
A stone is thrown at an angle θ to the horizontal reaches a maximum height H. Then the time of flight of stone will be
- (a)
\(\sqrt{2H\over g}\)
- (b)
2\(\sqrt{2H\over g}\)
- (c)
\(2\sqrt{2Hsin\theta}\over g\)
- (d)
\(\sqrt{2Hsin\theta}\over g\)
\(h={u^2sinθ\over g}\ and\ T={2usinθ\over g}\ or\ T^2={4u^2sin^2θ\over g^2}\)
\(∴ {T^2\over H}=\left(2usin\theta\over g\right)^2\times{2g\over u^2sin^2θ}={8\over g}\)
or \(T^2={8H\over g}\)
or \(T=2\sqrt{2H\over g }\)
If the range of a gun which fires a shell with muzzle speed v is R, then the angle of elevation of the gun is:
- (a)
\(cos^{-1}\left(v^2\over Rg\right)\)
- (b)
\(cos^{-1}\left(Rg\over v^2\right)\)
- (c)
\({1\over 2}sin^{-1}\left(v^2\over Rg\right)\)
- (d)
\({1\over 2}sin^{-1}\left( Rg\over v^2\right)\)
\(R={v^2sinθ\over g}\ or\ sin2θ={gR\over v^2}\)
\(∴\ θ={1\over 2}sin^{-1}\left(gR\over v^2\right)\)
In case of a projectile, where is the angular momentum minimum?
- (a)
At the starting point
- (b)
At the highest point
- (c)
On return to the ground
- (d)
At some location other than those mentioned above
A particle is projected upon an inclined plane with initial speed v = 20 m/s at an angle θ = 30° with the plane. The component of its velocity perpendicular to the plane when it strikes the plane is
- (a)
10√3 m/s
- (b)
10m/s
- (c)
5√3 m/s
- (d)
data is insufficient
Component of velocity ⊥ to plane remains the same (in opposite direction),
u sin θ = 20 sin 30°= 10m/s.
A particle of mass m is describing a circular path of radius r with uniform speed. If L is the angular momentum of the particle about the axis of the circle, the kinetic energy of the particle is given by:
- (a)
L2/mr2
- (b)
L22/mr2
- (c)
2L2/mr2
- (d)
mr2L
In circular motion of a particle of mass m along a path of radius r with a constant speed v (say), linear momentum = mu which is always ⊥ to the radius vector. Hence, angular momentum of the particle about the axis of the circle
= moment of linear momentum
= linear momentum x ..1distance W.r.t. axis of rotation
\(L = mvr,\ \therefore\ v={L\over mr}\) and \(K={1\over 2}mv^2={L^2\over 2mr^2}\)
Two particles of equal masses are revolving in circular paths of radii r1 and r2 respectively with the same period. The ratio of their centripetal force is:
- (a)
r1/r2
- (b)
\(\sqrt{r_1/r_2}\)
- (c)
(r1/r2)2
- (d)
(r2/r1)2
As T1=T2
Hence, \({2\pi r_1\over v1}={2\pi r^2\over v^2}\ or\ {v_1\over v2}={r1\over r_2}\)
\({F_1\over F_2}={mv_1^2\over r_1}\times{r_2\over mv_2^2}=\left(v_1\over v_2\right)^2\times{r2\over r_1}=\left(r_1\over r_2\right)^2\times{r_2\over r_1}={r_1\over r_2}\)
A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 see, it rotates through an angle θ1; in the next 2 see it rotates through an additional angle θ2. The ratio of \(θ_2\over θ_1\) is
- (a)
1
- (b)
2
- (c)
3
- (d)
5
\(θ=ω_0t+{1\over 2}\alpha t^2\)
Hence, \(ω_0=0, θ_1={1\over 2}\alpha(2)^2=2\alpha\)
\(θ_2={1\over 2}\alpha(4)^2-θ_1=8\alpha-2\alpha=5\alpha\)
\({θ_2\over θ1}={6\alpha\over 2\alpha}=3\)
A train is moving towards north. At one place it turns towards north-east. Here, we observe that
- (a)
the radius of curvature of outer rail will be greater than that of the inner rail
- (b)
the radius of curvature of the inner rail will be greater than that of outer rail
- (c)
the radius of curvature of one of the rails will be greater
- (d)
the radius of curvature of the outer and inner rails will be the same
Because the train turns towards north-east, hence the radius of curvature of outer rail will be greater than that of inner rail.
A pendulum consist mg (If a small sphere of mass M suspended by an inextensible and massless string of length t is made to swing in a vertical plane. If the breaking strength of the string is 2Mg, then the maximum angular amplitude of the displacement from the vertical can be:
- (a)
0°
- (b)
30°
- (c)
60°
- (d)
90°
A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :
- (a)
\(\pi{v^2\over g}\)
- (b)
\(\pi{v^4\over g^2}\)
- (c)
\({\pi\over2}{v^4\over g^2}\)
- (d)
\(\pi{v^2\over g^2}\)
\(R_{max}={v^2sin90^0\over g}={v^2\over g}\)
Area = \(\pi(R_{max})^2={\pi v^4\over g^2}\)
A particle of mass m is circulating on a circle of radius r having angular momentum L, then the centripetal force will be:
- (a)
L2/mr
- (b)
L2 m/r
- (c)
L2/mr3
- (d)
L2/mr2
Angular momentum, L = r x p = r X m x v
or \(v={L\over mr}\)...(i)
Now, as centripetal force \(F_C={mv^2\over r}\) ....(ii)
Substituting the value ofv from eqn. (i) in eqn. (ii), we get
\(F_c={m\over r}\left[L\over mr\right]^2={L^2\over mr^3}\)
Since, in the given case vcm is zero so its linear momentum will also be zero.
A bottle of soda water is held by the neck and swing briskly ill a vertical circle. Near which position of the bottle do the bubbles collect?
- (a)
Near the bottom
- (b)
Near the neck
- (c)
In the middle of the bottle
- (d)
Bubbles remain uniformly distributed
The earth (mass = 6 x 1024kg) revolves around the sun with an angular velocity of 2 x 10-7 radian/see in a circular orbit of radius 1.5 x 108 km. The force exerted by the sun, on the earth is:
- (a)
6x1019 N
- (b)
18x1025 N
- (c)
36x1021N
- (d)
27x1039N
M= 6x1024 kg, co = 2x10-7 rad/sec
r= 1.5x108 km = 1.5 x 1011m
Force exerted on the earth
= M\(\omega\)2r = (6 x 1024) x (2 X 10-7)2 x (1.5 X 1011)
=36x1021 N.
Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection, of one is \(\pi\)/3 and the maximum height reached by it is 102 m. Then the maximum height reached by the other (in metre) is:
- (a)
336
- (b)
224
- (c)
56
- (d)
34
Horizontal ranges are same for complementary angles of projection, i. e. , for \(\theta\) and \((90^0-\theta)\)
We know that if two stones have same horizontal range, then this implies that both are projected at \(\theta\) and \((90^0-\theta)\)
Here, \(\theta={\pi\over 3}=60^0\)
90° - \(\theta\) = 90° - 60° = 30°
For first stone:
Max. height = 102 = \(u^2sin^260^0\over 2g\)
For second stone:
Max. height = \({u^2sin^230^0\over 2g}\)
\(\therefore\ {h\over 102}={sin^230^0\over sin^260^0}={(1/2)^2\over (\sqrt3/2)^2}\)
\(h=102\times{1/4\over 3/4}=34\)
A body is suspended from a smooth horizontal nail by a string of length 0.25 m. What minimum horizontal velocity should be given to it in the lowest position, so that it may move in a complete vertical circle with the nail at the centre?
- (a)
\(\sqrt{12.25}ms^{-1}\)
- (b)
4.9ms-1
- (c)
\(7\sqrt2ms^{-1}\)
- (d)
\(\sqrt{9.8}ms^{-1}\)
The direction of motion of body with the horizontal at this instant is:
- (a)
tan-1 (2)
- (b)
tan-1 (1/2)
- (c)
45°
- (d)
0°
Two bodies are projected from ground with equal speeds 20 rnIs from the same position in same vertical plane to have equal range but at different angle above the horizontal. If one of the angle is 30°, the sum of their maximum heights is:
- (a)
400 m
- (b)
20 m
- (c)
30 m
- (d)
40 m
A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?
- (a)
2s
- (b)
4s
- (c)
1s
- (d)
6s
A particle moves in a circle of radius 30 cm. Its linear speed is given by v = 2t, where t in second and u in m/s. Find out its radial and tangential acceleration at t = 3 sec respectively:
- (a)
220 m/sec2, 50 m/sec2
- (b)
100 m/sec2, 5 m/sec2
- (c)
120 m/sec2, 2 m/sec2
- (d)
110 m/sec2, 10 m/sec2
Given that;
r = 30 cm = 0.3 m and v = 2t
∴ Radial acceleration at I = 3sec
\(a_r={v^2\over r}={4t^2\over 0.3}={4\times3^2\over 0.3}=120m/s^2\)
and tangential acceleration
\(a_t={dv\over dt}=2m/sec^2\)
A particle is describing uniform circular motion. Its acceleration is:
- (a)
along the radius of circular path pointing towards the centre
- (b)
along the tangent to the circular path
- (c)
along the radius of the circular path pointing away from the centre
- (d)
zero
When a particle is describing a uniform circular motion, its acceleration is along the radius and directed towards the centre. This acceleration is known as centripetal acceleration.
A person takes an aim at a monkey sitting on a tree and fires a bullet. Seeing the smoke the monkey begins to fall freely; then the bullet will:
- (a)
hit the monkey
- (b)
go above the monkey
- (c)
go below the monkey
- (d)
hit the monkey if the initial velocity of the bullet is more than a definite velocity