Absolute zero is obtained from

P–V graph
P-¹⁄_Vgraph
P–T graph
V–T graph

Solution

P α T if V is constant, where P is pressure of certain amount of gas and T is absolute zero temperature.

The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be at

– 6 V
+ 12 V
zero
– 12 V

Solution

When negative terminal is grounded, positive terminal of battery is at +12 V. When positive terminal is grounded, the negative terminal will beat –12 V.

The approximate value of quantum number n for the circular or bit of hydrogen 0.0001 mm in diameter is

1000
60
10000
31

Solution

A ray of light passes through an equilateral prism(μ= 1.5).The angle of minimum deviation is

45º
37º 12'
20º
30º

Solution

Relative density of a metal may be found with the help of spring balance. In air the spring balance reads (5.00 ± 0.05) N and in water it reads (4.00 ± 0.05) N. Relative density would be

(5.00 ± 0.05)N
(5.00 ± 11%)
(5.00 ± 0.10)
(5.00 ± 6%)

Solution

Relative density =\(\frac{Weight\, of\, body\, in\, air}{Loss\, of\, weight\, in\, water}\)

\(\frac{5.00}{5.00–4.00}=\frac{5.00}{1.00}\)

\(\frac{\Delta \rho }{\rho }\times 100=\left ( \frac{0.05}{5.00}+\frac{0.05}{1.00} \right )\times 100\)

= (0.01 +0.05) × 100

= 0.06 × 100 = 6%

∴Relative density =5.00±6%

In Rutherford scattering experiment, the number of ∝-particles scattered at 60° is 5 × 10⁶. The number of ∝-particles scattered at 120° will be

15 × 10⁶
35 × 10⁶
59 × 10⁶
None of these

Solution

Fusion reaction occurs at temperatures of the order of

10³K
10⁷ K
10 K
10⁴ K

The dimensional formula for relative density is

[ML^-3]
[M⁰L^-3]
[M⁰L⁰T^-1]
[M⁰L⁰T⁰]

A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 watt of power, the force exerted by it on the surface is

3.2 × 10^–8 N
3.2 × 10^–7 N
5.12 × 10^–7 N
5.12 × 10^–8 N

Solution

When 97.52 is divided by 2.54, the correct result is

38.3937
38.394
38.39
38.4

Solution

\(\frac{97.52}{2.54}\)=38.393=38.4(with least number of significant figures,3).

In the fig. S1 and S2 are identical springs. The oscillation frequency of the mass m is f. If one spring is removed, the frequency will become

f
f × 2
f × √2
f/√2

Solution

Two different metals are joined end to end. One end is kept at constant temperature and other end is heated to a very high temperature. The graph depicting the thermo e.m.f.is

Solution

At the heighest point O

E = maximum and A shows temperature of inversion at which emf changes in sign.

The thermo e.m.f. of a thermocouple is 25μ V/ºC at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as 10^–5A,is connected with the thermocouple. The smallest temperature difference that can be detected by this system is

12ºC
0ºC
20ºC
16ºC

Solution

A wire loop is rotated in a uniform magnetic field about an ax is perpendicular to the field. The direction of the current induced in the loop reverses once each

quarter revolution
half revolution
full revolution
two revolutions

Solution

It is because after every 1/2 revolution the current becomes zero and mode of change in flux changes there after (If before the current becomes zero, the mode of flux change was from left to right then after the current becomes zero the mode of flux change becomes right to left).

A pendulum clock is 5 seconds fast at temperature of 15ºC and 10 seconds slow at a temperature of 30ºC. At what temperature does it give the correct time? (take time interval =24 hours)

18ºC
20ºC
22ºC
25ºC

Solution

Δt = ¹⁄₂ α ΔT × t

∴ 5 = ¹⁄₂ α (T - 15) × 86400

and 10 = ¹⁄₂ α (30 - T) × 86400

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