If the ratio of lengths, radii and Young’s modulus of steel and brass wires shown in the figure are a, b, and c, respectively. The ratio between the increase in lengths of brass and steel wires would be

\(\frac{b^{2}a}{2c}\)
\(\frac{bc}{2a^{2}}\)
\(\frac{ba^{2}}{2c}\)
\(\frac{a}{2b^{2}c}\)

Solution

A horizontal overhead powerline is at height of 4m from the ground and carries a current of 100 A from east to west.The magnetic field directly below it on the ground is (μ₀ = 4π × 10^–7 Tm A^–1)

2.5 × 10^–7 T southward
5 × 10^–6 T northward
5 × 10^–6 T southward
2.5 × 10^–7 T northward

Solution

A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will be

V/2
\(V/\sqrt{2}\)
2 V
4 V

Solution

Since V_rms is doubled by increasing the temp. so by

\(V_{rms}=\sqrt{\frac{3KT}{m}}\), the temp. increase by four times.

Now for constant pressure \(\frac{V_{1}}{T_{1}}\)

V₁ = V, T₁ = T°K, T₂ = 4T°K, V₂ = ?.

V₂ = 4V

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

Statement 1 :In Lyman series, the ratio of minimum and maximum wavelength is ³⁄₄.

Statement 2 :Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom.

(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

(d)Statement-1 is False, Statement -2 is True

A force is given by F = at + bt², where t is time, the dimensions of a and b are

[M L T^–4] and [M L T^-1]
[M L T^-1] and [M L T⁰]
[M L T^–3] and [M L T^–4]
[M L T^–3] and [M L T⁰]

Solution

[at] = [F] amd [bt²] = [F]

⇒[a] = MLT^–3 and [b] = M L T^–4

A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m/s. To givean initial upward acceleration of 20 m/s²,the amount of gas ejected per second to supply the needed thrust will be(Take g= 10 m/s²)

127.5 kg/s
137.5 kg/s
155.5 kg/s
187.5 kg/s

Solution

When heat is given to a gas in an isothermal change, the result will be

external work done
rise in temperature
increase in internal energy
external work done and also rise in temperature

Volume of one mole gas changes according to the V = a/T. If temperature change is ΔT, then work done will be

RΔT
-RΔT
\(\frac{R}{\gamma -1}\Delta T\)
R (γ– 1)ΔT

Solution

If the potential energy of a gas molecule is U = M/r⁶– N/r¹², M and N being positive constants, then the potential energy at equilibrium must be

zero
M²/4N
N²/4M
MN²/4

Solution

A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ. The frictional force

dissipates energy as heat.
decreases the rotational motion.
decreases the rotational and translational motion.
converts translational energy to rotational energy

Solution

Net work done by frictional force when drum rolls down without slipping is zero.

W_net = 0, W_trans. + W_rot.= 0

ΔK_trans.+ ΔK_rot. = 0

ΔK_trans= –ΔK_rot.

i.e., converts translation energy to rotational energy.

The binding energy per nucleon for \(_{1}^{2}\)H and \(_{2}^{4}\)H erespectively are 1.1 MeV and 7.1 MeV. The energy released in MeV when two \(_{1}^{2}\)H nuclei fuse to form \(_{2}^{4}\)He is

4.4
8.2
24
28.4

Solution

The chemical reaction of process is \(2_{1}^{2}H\rightarrow _{2}^{4}He\)

Energy released = 4 × (7.1) - 4(1.1) = 24 eV

The length of a given cylindrical wire is increased by 100%.Due to the consequent decrease in diameter the change in the resistance of the wire will be

100%
50%
300%
200%

Solution

Three equal resistors connected across a source of e.m.f.together dissipate 10 watt of power. What will be the power dissipated in watts if the same resistors are connected in parallel across the same source of e.m.f.?

10
¹⁰⁄₃
30
90

Solution

An X-ray tube is operated at 15 kV. Calculate the upper limit of the speed of the electrons striking the target.

7.26 × 10⁷m/s
7.62 × 10⁷m/s
7.62 × 10⁷cm/s
7.26 × 10⁹m/s

Solution

Two identical bar magnets are fixed with their centres at a distance d apart. A stationary charge Q is placed at Pin between the gap of the two magnets at a distance D from the centre O as shown in the figure.

The force on the charge Q is

directed perpendicular to the plane of paper
zero
directed along OP
directed along PO

Solution

Force on a charged particle is given by F= qvB. Here v= 0 and also resultant B is zero.

∴ Force =0

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