Physics Test 133

Result

Physics Test 133

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Question 1
4 / -1

A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the direction of magnetic field. The radius of the loop is made to shrink at a constant rate of 1 mm s^-1. Find the induced emf in the loop when it's radius is 2 cm.

A
2 πμV

B
πμV

C
1 μV

D
2 μV

Solution

Here;
Magnetic field. B = 0.025 T
Radius of the loop, r = 2 cm = 2 x 10^-2 m
Constant rate at which radius of the loop shrinks,
Question 2
4 / -1

A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.

A
-2.94 cm

B
-3.12 cm

C
2.24 cm

D
1.52 cm

Solution

Here, A = 5 cm ; f = 5 Hz ; t = 0.32 s
Now,y = A sin ωt = A sin 2π f t = 5 sin (576°) = –2.94 cm
Question 3
4 / -1

For network shown in figure, determine V₀and I_D. (Where, I_D = Current flowing in the diodes)

A
11 V; 1.96 mA

B
16 V; 2.32 mA

C
21 V; 2.84 mA

D
28 V; 3.49 mA

Solution

Direction of current l_D is the same direction as arrow in the diode symbol. Therefore, both the diodes are in forward Bias; hence voltage drops across their terminals are V_Si and V_Ge are 0.7 V and 0.3 V respectively.
Question 4
4 / -1

A particle is projected vertically upwards with a velocity u, from a point O. When it returns to the point of projection, which of the following is incorrect ?

A
Its average velocity is zero

B
Its displacement is zero

C
Its average speed is u

D
Its average speed is u/2

Solution

Total displacement is 0 as the particle returns to the original position, thus the average velocity zero.
Total distance travelled is 2s and total time taken is 2t
0² = u² - 2gs

Hence option (Its average speed is u) is correct.
Question 5
4 / -1

Two blocks each of mass m, lie on a smooth table. They are attached to two other masses as shown in figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces with respect to each other is

A
5g/6

B
5g/3

C
17g/12

D
17g/6

Solution

In a plane mirror the image moves twice as faster as mirror thus
acceleration of image in mirror AB = a → AB
acceleration of image in mirror CD= a → cd
acceleration of images w.r.t. each other is
Question 6
4 / -1

An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV. The kinetic energy is E and its de Broglie wavelength is λλ. Then

A
E = 6.8 eV, λ = 6.6 × 10⁻¹⁰m

B
E = 3.4 eV, λ = 6.6 × 10⁻¹⁰m

C
E = 3.4 eV, λ = 6.6 × 10⁻¹¹m

D
E = 6.8 eV, λ = 6.6 × 10⁻¹¹m

Solution
Question 7
4 / -1

Two cylindrical rods of uniform cross-sectional area A and 2A, having free electrons per unit volume 2n and n respectively are joined in series. A constant current I flows through them in steady state. The ratio of drift velocity of free electrons in the left rod to that of the right rod is (VL/VR) is:

A
1/2

B
1

C
2

D
4

Solution

Since current I = neAV_d through both rods is same
2(n)e AV_L= ne (2A) V_R
or V_L/V_R=1
Question 8
4 / -1

Uncharged capacitor of capacitance 4 μF4 and a resistance of 2.5 M ΩM Ω are connected in series with 12 V battery at t=0. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given, ln (2) = 0.693]

A
13.86 s

B
6.93 s

C
20 s

D
14 s

Solution

Given : V_C = 3VR = 3(V - V_C)
Here: V is the applied potential.

⇒ t = 2 x 10 x 0.693 = 13.86 sec
Question 9
4 / -1

A refrigerator is to maintain the eatables, kept inside it, at 9^oC. The coefficient of performance of the refrigerator, if the room temperature is 36^oC, is

A
10.45

B
4

C
3

D
11.45

Solution

Here, T₁ = 36°C = 36 4 - 273 = 309 K
T₂ = 9°C = 9 + 273 = 282 K
The coefficient of performance of a refrigerator is given by
Question 10
4 / -1

A ball is projected up an incline of 30o with a velocity of 30 ms⁻¹ at an angle of 30^o with reference to the inclined plane from the bottom of the inclined plane. If g = 10 ms⁻², then the range on the inclined plane is

A
12 m

B
60 m

C
120 m

D
600 m

Solution