Physics Test 211

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Physics Test 211

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Question 1
4 / -1

The power factor of an ac circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is

A
R/ωL

B
R/(R² + ω²L²)^1/2

C
ωL/R

D
R/(R² - ω²L²)^1/2

Solution
Question 2
4 / -1

Two point charges, q and 4q are held a distance 'r' apart. The electric field due to them is zero at a distance

A
r/√3 from charge 4q

B
r/3 from charge 4q

C
2r/√3 from charge 4q

D
2r/3 from charge 4q

Solution

Let the electric field be zero at a distance x from charge 4q. Then
Question 3
4 / -1

What should be the angle of banking of the road if a vehicle, travelling with speed v, negotiates a curve of radius r?

A
tan θ = v²/2

B
tan θ = v²/rg

C
tan θ = v./2g

D
None of these

Solution

On the banked road the component of the normal reaction N cos θ will balance the weight of the vehicle

N cos θ = mg …..1
Question 4
4 / -1

Let there be a spherically symmetric charge distribution with charge density varying as where r is the distance from the origin. The electric field at a distance r(rltR) from the origin is given by

A

B

C

D

Solution

Let us consider a spherical shell of radius x and thickness dx.

Charge on this shell

∴ Total charge in the spherical region from centre to r (rltR) is
Question 5
4 / -1

Electromagnetic waves are transverse in nature, is evident by

A
polarisation

B
interference

C
reflection

D
diffraction

Solution

Interference and diffraction explain the wave nature of light or EM waves. Polarisation is the phenomenon by which we restrict the vibration of wave in a particular direction perpendicular to direction of wave propagation. So, it explains the transverse nature of EM waves.
Question 6
4 / -1

A car is moving on a straight road. If its engine is delivering constant power, then the distance covered by the car in time t is proportional to

A
t^1/2

B
t^2/3

C
t^3/2

D
t²

Solution

Power is defined as the dot product of force and velocity.

As the car is moving in a straight line with constant power, the direction of the force and velocity will be same.

So, P = Fv and F = ma

So, P = mav

Acceleration of the car over the time interval 't' starting from rest is given by:
Question 7
4 / -1

A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then, the true statement about the scale reading is

A
both the scales read M kg each

B
the scale of the lower one reads M kg and of the upper one zero

C
the reading of the two scales can be anything, but the sum of the reading will be M kg

D
both the scales read M/2 kg

Solution

Hence, both scales will show the same reading.
Question 8
4 / -1

Two particles starting from a point on a circle of radius 4 m in a horizontal plane move along the circle in opposite directions with constant speeds of 4 ms^-1 and 6 ms^-1. The particles will collide with each other in

A
3.0 s

B
2.5 s

C
2.0 s

D
1.5 s

Solution

As the particles are moving in opposite directions,

Relative speed of the particles = 6 + 4 = 10 ms^-1

Distance covered = 2πR = 2 × 3.14 × 4
Question 9
4 / -1

Which of the following represents the Boolean function for the given circuit of gates?

A
x'y + x'z + y'z

B
x'y + xz + y'

C
x' + x'z' + y'z

D
x + x'z + y'

Solution
Question 10
4 / -1

When a system is taken from state i to state f along the path iaf, it is found that Q = 50 cal and W = 20 cal. Along the path ibf, Q = 36 cal. W along the path ibf is

A
14 cal

B
6 cal

C
16 cal

D
66 cal

Solution

According to first law of thermodynamics, for the path iaf,

Q_iaf = ΔU_iaf + W_iaf

or, ΔU_iaf = Q_iaf - W_iaf

= 50 - 20 = 30 cal

For the path ibf,

Q_ibf = ΔU_iaf + W_ibf

Since ΔU_iaf = ΔU_ibf, change in internal energy is path independent.

Q_ibf = ΔU_iaf + W_iaf

∴ W_ibf = Q_ibf - ΔU_iaf= 36 - 30 = 6 cal