The dimensions of universal gas constant are

[L² M¹ T^-2 K^-1]
[L¹ M² T^-2 K^-1]
[L¹ M¹ T^-2 K^-1]
[L² M² T^-2 K^-1]

Solution

\(R=\frac{PV}{\mu T}=\frac{W}{\mu T}=\frac{ML^{2}T^{-2}}{mol\, K}\)

where μ is number of mole of the gas

= [M¹L²T^-2K^-1mol^-1]

A closed pipe and an open pipe have their first overtones identical in frequency.Their lengths are in the ratio

1 : 2
2 : 3
3 : 4
4 : 5

Directions: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-

Statement 1 :If the temperature of a semiconductor is increased then it’s resistance decreases.

Statement 2 :The energy gap between conduction ban d and valence band is very small.

(a)Statement -1 is true, Statement-2 is true;Statement -2 is a correct explanation for Statement-1
(b)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is false
(d)Statement -1 is false, Statement-2 is true

Solution

(a)Statement -1 is true, Statement-2 is true;Statement -2 is a correct explanation for Statement-1

In semiconductors the energy gap between conduction band and valence band is small (≈1 eV). Due to temperature rise, electron in the valence band gain thermal energy and may jumpy across the small energy gap, (to the conduction band). Thus conductivity increases and hence resistance decreases.

The load versus elongation graph for four wires is shown.The thinnest wire is

Rectilinear motion of light in a medium is caused due to

high frequency
short wavelength
velocity of light
uniform refractive index of the medium

Solution

If the medium is heterogeneous having a gradient of refractive index. Then light rays will not follow a rectilinear(straight line path).

Two drops of the same radius are falling through air with asteady velocity of 5 cm per sec. If the two drops coalesce,the terminal velocity would be

10 cm per sec
2.5 cm per sec
5 × (4)^1/3 cm per sec
5 × √3 cm per sec

Solution

Angular momentum is

a polar vector
an axial vector
a scalar
None of these

Solution

Angular momentum \(\dot{L}\) is defined as \(\dot{L}=\dot{r}\times m(\dot{v})\) so \(\dot{L}\) is, an axial vector..

Electrons in the atom are held to the nucleus by

coulomb’s force
nuclear force
vander waal’s force
gravitational force

Which of the following must be known in order to determine the power output of an automobile?

Final velocity and height
Mass and amount of work performed
Force exerted and distance of motion
Work performed and elapsed time of work

Solution

Power is defined as the rate of doing work.For the automobile, the power output is the amount of work done (overcoming friction)divided by the length of time in which the work was done.

The velocity of projection of a body is increased by 2%.Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

Solution

A rod PQ of mass M and length L is hinged at end P.The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

g/L
2g/L
^2g⁄_3L
^3g⁄_2L

Solution

M_p denotes the mass of a proton and M_n that of a neutron.A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light)

M(N, Z) = NM_n + ZM_p + B/c²
M(N, Z) = NM_n + ZM_p – Bc²
M(N, Z) = NM_n + ZM_p + Bc²
M(N, Z) = NM_n + ZM_p – B/c²

Solution

Iron is ferromagnetic

at all temperatures
at NTP only
above 770°C
below 770°C

The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167WK^-1m^-1 and thickness1 cm. The temperature is maintained by circulating oil as shown in figure. Find the radiation loss to the surrounding in Jm^-2s^-1 if temperature of the upper surface of the disc is 27°C and temperature of the surrounding is 27°C

595 Jm^-2s^-1
545 Jm^-2s^-1
495 Jm^-2s^-1
None of these

Solution

The rate of heat loss per unit area due to radiation

= ∈σ(T⁴–T₀⁴)

= 0.6 × 5.67 × 10^-8[(400)⁴–(300)⁴] = 595 Jm^–2s^-1
.

The dimensional formula for relative refractive index is

[M⁰ L¹ T^-1]
[M⁰ L⁰ T⁰]
[M⁰ L¹ T¹]
[MLT^-1]

Solution

\(\mu _{r}=\frac{\mu }{\mu _{0}}=\left [ M^{0}L^{0}T^{0} \right ]\)

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