Physics Test 187

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Physics Test 187

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Weekly Quiz Competition

Question 1
4 / -1

A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be

A
1 rad/sec²

B
0.25 rad/sec²

C
0.5 rad/sec²

D
2 rad/sec²

Solution

F = ma

a = Wr

So, F = mwr

a = wr = acceleration

m = 200kg

F = 100 n-m, r = 2m, w = to be calculated

100 = 200 x w²

w = 0.25 rad/sec²
Question 2
4 / -1

A solid sphere and a solid cylinder of same mass are rolled down on two inclined planes of heights h₁ and h₂ respectively. If at the bottom of the plane the two objects have same linear velocities, then the ratio of h₁ : h₂ is

A
2 : 3

B
7 : 5

C
14 : 15

D
15 : 14

Solution

for solid sphere and 1/2 for solid cylinder At the bottom, both have the same linear velocities, i.e., they have the same translational kinetic energies
Question 3
4 / -1

A particle of mass m moves along line PC with velocity ν as shown. What is the angular momentum of the particle about P?

A
mvL

B
mvl

C
mvr

D
zero

Solution

The particle moves with linear velocity vv along line PC. The line of motion is through P. Hence angular momentum is zero.
Question 4
4 / -1

A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momentum is

A
L/4

B
L/2

C
L

D
2L

Solution

L = mr² ω. For given m and ω, L ∝ r². If r is halved, the angular momentum L becomes one-fourth.
Question 5
4 / -1

Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

A
0.42 m from mass of 0.3 kg

B
0.70 m from mass of 0.7 kg

C
0.98 m from mass of 0.3 kg

D
0.98 m from mass of 0.7 kg

Solution

Work done W = 1/2Iω²

if x is the distance of mass 0.3 kg from the centre of mass, we will have,
Question 6
4 / -1

Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2 m.

A
0

B
1

C
2

D
3

Solution

Force of gravity acts on the car vertically downward while the car is moving along the horizontal road, i.e., the angle between them is 90°.

Work done by the car against gravity,

W = Fscos90° = 0
Question 7
4 / -1

The potential energy for a conservative system is given by U = ax² – bx, where aa and bb are positive constants. The law of the force governing the system is

A
F = constant

B
F = bx – 2a

C
F = b – 2ax

D
F = 2ax

Solution

In a conservative field F = -dU / dr

∴ F = −d / dx(ax²− bx) = b − 2ax
Question 8
4 / -1

The average energy consumed by a human being in a day is

A
2400 cal

B
2400 J

C
2400 kJ

D
2400 kcal

Solution

The average energy consumed by a human being in a day is
Question 9
4 / -1

The kilowatt-hour is the unit of

A
Time

B
Power

C
Energy

D
Force

Solution

Kilowatt is the unit of power and hour is the unit of time,

The basic definition of power is rate of change of work done or energy with respect to time.

Power × Time = Energy

Kilowatt-hour is the unit of energy.
Question 10
4 / -1

For inelastic collision between two spherical rigid bodies

A
The total kinetic energy is conserved

B
The total mechanical energy is not conserved

C
The linear momentum is not conserved

D
The linear momentum is conserved

Solution

In an inelastic collision, the particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus the kinetic energy of particles no longer remains conserved. However, in the absence of external forces, law of conservation of linear momentum still holds good.

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Re-Attempt Weekly Quiz Competition

Physics Test 187

Result

Physics Test 187

Score

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Rank

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SHARING IS CARING

Question 1

1 rad/sec2

0.25 rad/sec2

0.5 rad/sec2

2 rad/sec2

Solution

Question 2

2 : 3

7 : 5

14 : 15

15 : 14

Solution

Question 3

mvL

mvl

mvr

zero

Solution

Question 4

L/4

L/2

L

2L

Solution

Question 5

0.42 m from mass of 0.3 kg

0.70 m from mass of 0.7 kg

0.98 m from mass of 0.3 kg

0.98 m from mass of 0.7 kg

Solution

Question 6

0

1

2

3

Solution

Question 7

F = constant

F = bx – 2a

F = b – 2ax

F = 2ax

Solution

Question 8

2400 cal

2400 J

2400 kJ

2400 kcal

Solution

Question 9

Time

Power

Energy

Force

Solution

Question 10

The total kinetic energy is conserved

The total mechanical energy is not conserved

The linear momentum is not conserved

The linear momentum is conserved

Solution

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1 rad/sec²

0.25 rad/sec²

0.5 rad/sec²

2 rad/sec²