Two electric bulbs whose resistance are in the ratio 1 : 2 are arranged in parallel to a constant voltage source. The powers dissipated in them have the ratio

1 : 2
1 : 1
2 : 1
1 : 4

A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving northward with the same speed v. The two particles coalesce on collision. The new particle of mass 2m will move in the north-external direction with a velocity :

v/2
2v
v /√2
None of these

Solution

There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joint connections is R₀. The net resistance of the whole grid between the points A and B as shown is

R₀
^R₀⁄₂
^R₀⁄₃
^R₀⁄₄

Solution

The ratio of work function and temperature of two emitters are 1 :2, then the ratio of current densities obtained by them will be

4 : 1
2 : 1
1 : 2
1 : 4

Solution

A wire of resistance 4 Wis stretched to twice its original length. The resistance of stretched wire would be

4 Ω
8 Ω
16 Ω
2 Ω

Solution

A uniform rope of length L resting on a friction less horizontal surface is pulled at one end by a force F. What is the tension in the rope at a distance ι from the end where the force is applied.

F
F (1 + ι/L)
F/2
F (1 – ι/L)

Solution

Let n be the mass per unit length of rope. Therefore,mass of rope = nL.

Acceleration in the rope due to force F will be a = F/nL.

Mass of rope of length (L– ι) will be n (L – ι).

Therefore, tension in the rope of length (L – ι), is equal to pulling force on it

= n (L – ι) a =n (L – ι) × F/nL =F (1 – ι/L)

The time taken by the current to rise to 0.63 of its maximum value in a d.c. circuit containing inductance (L)and resistance (R) depends on

L only
R only
^L⁄_R
LR

A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, what is the force on the arm AC?

\(-\sqrt{2}\overrightarrow{F}\)
\(-\overrightarrow{F}\)
\(\overrightarrow{F}\)
\(\sqrt{2}\overrightarrow{F}\)

Solution

In the equation X = 3YZ², X and Z are dimensions of capacitance and magnetic induction respectively. In MKSQ system, the dimensional formula for Y is

[M^-3 L^-2 T^-2 Q^-4]
[M L^-2]
[M^-3 L^-2 Q⁴ T⁸]
[M^-3 L^-2 Q⁴ T⁴]

Solution

[Y]=\(\frac{[X]}{[Z^{2}]}=\frac{M^{-1}L^{-2}T^{4}A^{2}}{M^{2}T^{-4}A^{-2}}\)=M^-3L^-2Q⁴T⁴

\(\left ( A=\frac{Q}{T} \right )\)

In Young’s double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is

sin^–1(0.01)
sin^–1(0.0001)
sin^–1(0.001)
sin^–1(0.1)

Solution

The displacement x of a particle varies with time t as x = ae^-αt+ be^βt, where a, b, α and β are positive constants.The velocity of the particle will

be independent of α and β
drop to zero when α = β
go on decreasing with time
go on increasing with time

Solution

The time period of the oscillating system (see figure) is

\(T=2\pi \sqrt{\frac{m}{k_{1}k_{2}}}\)
\(T=2\pi \sqrt{\frac{m}{k_{1}+k_{2}}}\)
\(T=2\pi \sqrt{mk_{1}+k_{2}}\)
None of these

Two vibrating tuning forks produce progressive waves given by Y₁= 4 sin 500 pt and Y₂= 2 sin 506 pt. Number of beats produced per minute is

Solution

Equation of progressive wave is given by

Y = A sin 2πf t

Given Y₁= 4 sin 500 πt and Y₂= 2 sin 506πt.

Comparing the given equations with equation of

progressive wave, we get

2f₁= 500,⇒ f₁ = 250

2f₂= 506 ⇒ f₂ = 253

Beats = f₂– f₁= 253 – 250 = 3 beats/sec

= 3 × 60 = 180 beats/minute.

A ray of light traveling in water is incident on its surface open to air. The angle of incidence is q, which is less than the critical angle. Then there will be

only a reflected ray and no refracted ray
only a refracted ray and no reflected ray
a reflected ray and a refracted ray and the angle between them would be less than 180°–2q
are flected ray and a refracted ray and the angle between them would be greater than 180°–2q

Solution

The ray is partly reflected and partly refracted.

∠MOB = 180 – 2θ

But the angle between refracted and reflected ray is

∠POB. Clearly ∠POB is less than ∠MOB.

A vessel contains air at a temperature of 15º C and 60 % R.H.What will be the R.H. if it is heated to 20º C? (S.V.P. at 15º C is 12.67 & at 20º C is 17.36 mm of Hg respectively)

262 %
26.2 %
44.5 %
46.2 %

Solution

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