Work done by a conservative force is positive if

P.E. of the body increases
P.E. of the body decreases
K.E. of the body increases
K.E. of the body decreases

A steel rod of radius R = 10mm and length L= 100 cm is stretched along its length by a force F = 6.28 ×10⁴ N. If the Young’s modulus of steel is Y = 2 × 10¹¹ N/m²,the percentage elongation in the length of the rod is :

0.100
0.314
2.015
1.549

Solution

A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has as frequency 2f. If v be the velocity of sound,then the velocity of the car, in the same velocity units, will be

v/2
v/√2
v/3
v/4

Solution

One face of a rectangular glass plate 6 cm thick is silvered. An object held 8cm in front of the first face, forms an image 12 cm behind the silvered face. The refractive index of the glass is

Solution

Unit of latent heat is

N Kg^–1
J mol^–1
J Kg^–1
N mol^–1

Solution

\(L = \frac{Q}{m} = \frac{J}{kg} = Jkg^{-1}\)

A person travels along a straight road for the first half time with a velocity v1 and the second half time with a velocity v2. Then the mean velocity v is given by

\(\overline{V}=\frac{V_{1}+V_{2}}{2}\)
\(\frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}\)
\(\overline{v}=\sqrt{v_{1}v_{2}}\)
\(\overline{v}=\sqrt{\frac{V_{2}}{V_{1}}}\)

Solution

Let for the first half time t, the person travels a distance

s₁. Hence \(v_{1}=\frac{s_{1}}{t}\: or\: s_{1}=v_{1}t\)

For second half time,\(v_{2}=\frac{s_{2}}{t}\: or\: s_{2}=v_{2}\: t\)

Now, \(\overline{v}=\frac{Total\: displacement}{Total\: time}=\frac{s_{1}+s_{2}}{2t}\)

\(=\frac{v_{1}t+v_{2}t}{2t}=\frac{v_{1}+v_{2}}{2}\)

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation P =kV. Molar heat capacity of the gas for the process is

C = C_v + R
C = C_v + ^R⁄₂
C = C_v – ^R⁄₂
C = C_v + 2R

Solution

A solid sphere of mass 1 kg rolls on a table with linear speed 1 ms^–1. Its total kinetic energy is

1 J
0.5 J
0.7 J
1.4 J

Solution

Graph of specific heat at constant volume for a monatomic gas is

Solution

For a monatomic gas

C_v = ³⁄₂ R

So correct graph is

Doubly ionised helium atom and hydrogen ions are accelerated, from rest, through the same potential difference.The ratio of final velocities of helium and hydrogen is

1 :√2
√2 : 1
1 : 2
2 : 1

Solution

The length of a metal is l₁ when the tension init is T₁ and is l₂ when the tension is T₂. The original length of the wire is

\(\frac{\iota _{1}+ \iota _{2}}{2}\)
\(\frac{\iota _{1}T_{2}+ \iota _{2}T_{1}}{T_{1}+T_{2}}\)
\(\frac{\iota _{1}T_{2}- \iota _{2}T_{1}}{T_{1}-T_{2}}\)
\(\sqrt{T_{1}T_{2}\iota _{1}\iota _{2}}\)

Solution

When a body is heated, which colour corresponds to high temperature

red
yellow
white
orange

Solution

When a body is heated then relation between colours and temperature is according to Prevost’s theory of radiation which states that everybody emitting radiant energy in all directions at a rate depending only on the nature of its surface and its temperature e.g., when a body is placed in an enclosure (furnace) it would acquire the temperature of furnace and seem white means radiate white light. So it becomes first dark and then white.

A monochromatic light wave of wave length λ enters a medium of refractive index μ= 2.Which of the following statements is true ?

speed and frequency are halved and wave length does not change
speed and length are halved and frequency does not change
frequency is frequency are halved while wave length and speed does not change
Wave length, speed and frequency are all halved

Solution

Frequency is the fundamental characteristic of the wave which does not change from one to other medium.

The X-rays of wavelength 0.5 Å are scattered by a target.What will be the energy of incident X-rays, if these are scattered at an angle of 72º ?

12.41 keV
6.2 keV
18.6 keV
24.82 keV

Solution

For an ideal gas graph is shown for three processes. Process1, 2 and 3 are respectively.

Isobaric, adiabatic, isochoric
Adiabatic, isobaric, isochoric
Isochoric, adiabatic, isobaric
Isochoric, isobaric, adiabatic

Solution

Isochoric proceess dV = 0

W = 0 proceess 1

Isobaric : W = PΔV = nRΔT

Adiabatic |W |= \(\frac{nR\Delta T}{\gamma -1}\) 0< γ– 1< 1 As work done in case of adiabatic process is more soprocess 3 is adiabatic and process 2 is isobaric.

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