Physics Test

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Physics Test - 17

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Question 1
1 / -0

The radius R of the soap bubble is doubled under isothermal condition. If T be the surface tension of the soap bubble, the work done in doing so is given by

A
32πR²T

B
24πR²T

C
8πR²T

D
4πR²T

Solution

The surface energy of a bubble

E = Tension × surface area

E=T×A

For soap bubble

E=2T×A

E=2T×4πR²

Here, R = radius of the bubble

If the radius is doubled then we have

E'=2T×4πR²

E'=2T×4π(2R)²

E'=32πR²T

Work done = E'−E

⇒ 32πR²T−8πR²T=24πR²T
Question 2
1 / -0

What is the change in the temperature on Fahrenheit scale and on Kelvin scale, if a iron piece is heated from 30 to 90^oC .

A
108^oF, 60K

B
100^oF, 55K

C
100^oF, 65K

D
60^oF, 108K

Solution

$Δ T_{C} = 90^{o} - 30^{o} C = 60^{o} C$

$Δ T_{F} = \frac{9}{5} Δ T_{C} = \frac{9}{5} (60^{o}) = 108^{o} F$

$Δ T = Δ T_{C} = 60 K$
Question 3
1 / -0

A knife of mass 1 kg slides on a frictionless wedge of mass 5 kg between two frictionless guides. At the instant shown in figure, the spring has initial compression 1m. Ground is also frictionless. Velocity of wedge when knife is about to touch the ground is (assume spring is ideal for large compressions) -

A
$\frac{6}{\sqrt{89}} m / s$

B
$\frac{12}{\sqrt{89}} m / s$

C
$\frac{2}{\sqrt{89}} m / s$

D
$\frac{24}{\sqrt{89}} m / s$

Solution

By conservation of mechanical energy

$1 \times 10 \times 3 + \frac{1}{2} \times 1 \times 12 = \frac{1}{2} \times 1 \times {(\frac{3}{4} v_{w})}^{2} + \frac{1}{2} \times 5 \times V_{W}^{2} + \frac{1}{2} \times 1 \times 5^{2}$

( v_w : velocity of wedge)

⇒ $v_{w} = \frac{24}{\sqrt{89}} m / s$
Question 4
1 / -0

The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, then the angle of inclination of the plane is

A
$\tan^{- 1} (\frac{3}{4})$

B
$\tan^{- 1} (\frac{1}{4})$

C
45°

D
30°

Solution

case (i)

F₁= Mg sinθ + fr

= Mg (sinθ + μ cos θ) ... (i)

Case (ii)

F₂+fr = Mg sinθ

F₂= Mg cosθ−μ Mg cosθ

= Mg (sinθ−μ cosθ) ... (ii)

F₁= 2F₂ (Given)

From Eqs. (i) and (ii)

Mg (sinθ+μ cosθ)=2Mg (sinθ−μ cosθ)

⟹ sinθ+1/4 cosθ=2(sinθ−1/4 cosθ)

⟹ cosθ(1/4 + 1/2) =sinθ

⟹ 3/4 cos θ=sin θ

⟹ tan θ=3/4

⟹ tanθ=3/4

⟹ θ=tan⁻¹(3/4)
Question 5
1 / -0

When the angle of incidence on the material is 60°, the reflected light is completely polarised. The velocity of refracted ray inside the material is

A
3×10⁸m/s

B
$\frac{3}{\sqrt{2}} \times 10^{8} m / s$

C
$\sqrt{3} \times 10^{8} m / s$

D
$\frac{1}{3} \times 10^{8} m / s$

Solution

$\mu=\tan i_{p}$

$\therefore \frac{c}{v}=\tan i_{p}$
$\therefore v=\frac{3 \times 10^{8}}{\sqrt{3}}=\sqrt{3} \times 10^{8} \mathrm{m} / \mathrm{s}$
Question 6
1 / -0

Force acting on a particle moving in a straight line varies with the velocity of the particle as $F = \frac{K}{v}$ . Here K is constant. The work done by this force in time t is

A
$\frac{K}{v^{2}} . t$

B
$2 K t$

C
$K t$

D
$\frac{2 K t}{v^{2}}$

Solution

According to the question :

∵ $F = \frac{K}{v}$

∴ $m \frac{d v}{d t} = \frac{K}{v} \Rightarrow v d v = \frac{K}{m} d t$

⇒ $\int v d v = \frac{K}{m} \int d t \Rightarrow \frac{v^{2}}{2} = \frac{K}{m} t$

⇒ $\frac{1}{2} m v^{2} = K t \Rightarrow W = K t$
Question 7
1 / -0

Water is filled in a cylindrical container to a height of 3 m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g=10 m s⁻²)

A
50 m² s⁻²

B
54 m² s⁻²

C
51 m² s⁻²

D
52 m² s^-2

Solution

Applying continuity equation at 1 and 2, we have

A₁V₁=A₂V₂ .....(i)

Further applying Bernoulli’s equation at these two points, we have

$\mathrm{P}_{0}+\mathrm{pgh}+\frac{1}{2} \mathrm{pv}_{1}^{2}=\mathrm{P}_{0}+0+\frac{1}{2} \mathrm{pv}_{2}^{2} \quad \ldots$(2)

Solving Eqs. (i) and (ii), we have

$\mathrm{v}_{2}^{2}=\frac{2 \mathrm{gh}}{1-\frac{\mathrm{A}_{2}^{2}}{\mathrm{A}_{1}^{2}}}$

Substituting the values, we have

$\mathrm{v}_{2}^{2}=\frac{2 \times 10 \times 2.475}{1-(0.1)^{2}}=50 \mathrm{m}^{2} \mathrm{s}^{-2}$
Question 8
1 / -0

If r be the distance of a point on the axis of a bar magnet from its centre, the magnetic field at this point is proportional to

A
$\frac{1}{r}$

B
$\frac{1}{r^{2}}$

C
$\frac{1}{r^{3}}$

D
$\frac{1}{r^{4}}$

Solution

Magnetic field at a distance r on the axis of the bar magnet is given as

$B=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 M}{r^{3}}$
$\therefore B \propto \frac{1}{r^{3}}$
Question 9
1 / -0

A single slit of width b is illuminated by a coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 cm from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)

A
4.5 cm

B
1.5 cm

C
6.0 cm

D
3.0 cm

Solution

For secondary minima dsinθ=nλ

sinθ= $\frac{n λ}{d}$

For second minima

n=2

$\sin θ = \frac{2 λ}{d} = \tan θ_{1} = \frac{x_{1}}{D}$

For fourth minima n=4

$\sin θ_{2} = \frac{4 λ}{d} = \frac{x_{2}}{D}$

$x_{2} - x_{1} = \frac{4 λ}{d} - \frac{2 λ}{d} = \frac{2 λ}{d} = 6 - 3 = 3 c m$

Width of central max= $\frac{2 λ}{d}$ =3 cm
Question 10
1 / -0

A point charge Q is placed at the corner of a square plate of side a then the flux through the square plate is-

A
$\frac{Q}{8 \in_{0}}$

B
$\frac{Q}{4 \in_{0}}$

C
$\frac{Q}{\in_{0}}$

D
$Z e r o$

Solution

The electric field due to Q at any point of square plate will be along the plane of square therefore angle between electric field and area vector of plate is 90°.∴ = 0

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Physics Test - 17

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Physics Test - 17

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SHARING IS CARING

Question 1

32πR2T

24πR2T

8πR2T

4πR2T

Solution

Question 2

108oF, 60K

100oF, 55K

100oF, 65K

60oF, 108K

Solution

Question 3

689m/s

1289m/s

289m/s

2489m/s

Solution

Question 4

tan−134

tan−114

45°

30°

Solution

Question 5

3×108m/s

32×108m/s

3×108m/s

13×108m/s

Solution

Question 6

Kv2.t

2Kt

Kt

2Ktv2

Solution

Question 7

50 m2 s−2

54 m2 s−2

51 m2 s−2

52 m2 s-2

Solution

Question 8

1r

1r2

1r3

1r4

Solution

Question 9

4.5 cm

1.5 cm

6.0 cm

3.0 cm

Solution

Question 10

Q8∈0

Q4∈0

Q∈0

Zero

Solution

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32πR²T

24πR²T

8πR²T

4πR²T

108^oF, 60K

100^oF, 55K

100^oF, 65K

60^oF, 108K

$\frac{6}{\sqrt{89}} m / s$

$\frac{12}{\sqrt{89}} m / s$

$\frac{2}{\sqrt{89}} m / s$

$\frac{24}{\sqrt{89}} m / s$

$\tan^{- 1} (\frac{3}{4})$

$\tan^{- 1} (\frac{1}{4})$

3×10⁸m/s

$\frac{3}{\sqrt{2}} \times 10^{8} m / s$

$\sqrt{3} \times 10^{8} m / s$

$\frac{1}{3} \times 10^{8} m / s$

$\frac{K}{v^{2}} . t$

$2 K t$

$K t$

$\frac{2 K t}{v^{2}}$

50 m² s⁻²

54 m² s⁻²

51 m² s⁻²

52 m² s^-2

$\frac{1}{r}$

$\frac{1}{r^{2}}$

$\frac{1}{r^{3}}$

$\frac{1}{r^{4}}$

$\frac{Q}{8 \in_{0}}$

$\frac{Q}{4 \in_{0}}$

$\frac{Q}{\in_{0}}$

$Z e r o$