Physics Test

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Physics Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

In an electron gun, the electrons are accelerated by the potential V. If e is the charge and m is the mass of an electron, then the maximum velocity of these electrons will be

A
2eV/m

B
√2eV/m

C
√2m/eV

D
V²/2em
Question 2
1 / -0

A bar magnet dropped into a coil of a conducting wire, along its axis, will fall with an acceleration

A
equal to g

B
less than g

C
more than g

D
first '1' then'3'
Question 3
1 / -0

A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is

A
3.67 A

B
2.67 A

C
1.67 A

D
2.40 A

Solution

Solution

Given : V1=4000 volts I1=100 A V2=240000 volts

Using V1I1=V2I2

∴ 4000×100=240000×I2 ⟹I2=1.67 A
Question 4
1 / -0

A 220 V, 1000 watt bulb is connected across a 110 V mains supply. The power consumed will be

A
250 W

B
500 W

C
1000 W

D
750 W
Question 5
1 / -0

Consider the situation in the figure. The work done in taking a point charge from P to Ais WA, from P to B is WB and from P to C is WC.

A
W _A < W_B < W_C

B
W_A = W_B= W_C

C
W_A > W_B > W_C

D
W_A = W_B < W_C

Solution

Solution

WA = WB = WC

Points A, B and C lie at the same distance from the charge q, i.e. they are lying on an equipotential surface. So, work done in moving a charge from A to B (WAB) or B toC (WBC) is zero.

Hence, work done in bringing a charge from P to A = WA,

from P to B, WB = WA+WAB = WA

and from P to C, WC = WA + WAB + WBC = WA

Hence, WA = WB = WC
Question 6
1 / -0

A dust particle of mass 10^-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19 C) the dust particle carries

A
1

B
10

C
100

D
1000
Question 7
1 / -0

A charge q is placed at the corner of a cube of side l. The electric flux passing through the cube is

A
q /∈₀

B
q /6∈₀

C
q/8∈₀

D
q/ 32∈₀

Solution

Solution

Eight cubes, each side l, are required to form a Gaussian surface

So that the charge q at the corner of a small cube appears at the centre of the bigger cube

According to Gauss's law, the electric flux through the bigger cubes = q ∈ 0

Hence the electric flux through the given small cube = q 8∈ 0
Question 8
1 / -0

A capacitor of capacity C₁ is charged upto V volt and then connected to an unchanged capacitor of capacity C₂. The potential difference across each capacitor will be

A
C₂ V/ C₁ + C₂

B
C₁V/ C₁ + C₂

C
1 + C₂/ C₁

D
(1 - C₂/ C₁)V

Solution

Solution

Charge on first capacitor = q₁ = C₁₁

Charge on second capacitor = q₂ = 0

When they are connected, in parallel the total charge

q = q₁ + q₂

∴ q = C₁₁

and capacitance, C = C₁ + C₂

Let V' be the common potential difference across each capacitaor, then q = CV'

∴ V' = q C = C1V/ C₁ + C₂
Question 9
1 / -0

Dielectric constant of a medium is unity. Its permittivity will be:

A
Zero

B
1

C
8.854 x 10^{− 12}C² N^{− 1} m^{− 2}

D
8.854 x 10¹²C² N^-1 m^-2
Question 10
1 / -0

If a current is passed in a spring, it

A
gets compressed

B
gets expanded

C
oscillates

D
remains unchanged

Solution

Solution

Two conductors will attract each other if they have current flowing in same direction. in a spring each loop has current flowing in same direction resulting in contraction of the spring.

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Re-Attempt Weekly Quiz Competition

Physics Test - 2

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Physics Test - 2

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SHARING IS CARING

Question 1

2eV/m

√2eV/m

√2m/eV

V2/2em

Question 2

equal to g

less than g

more than g

first '1' then'3'

Question 3

3.67 A

2.67 A

1.67 A

2.40 A

Solution

Question 4

250 W

500 W

1000 W

750 W

Question 5

W A < W B < W C

W A = W B = W C

W A > W B > W C

W A = W B < W C

Solution

Question 6

1

10

100

1000

Question 7

q /∈0

q /6∈0

q/8∈0

q/ 32∈0

Solution

Question 8

C2 V/ C1 + C2

C1V/ C1 + C2

1 + C2/ C1

(1 - C2/ C1)V

Solution

Question 9

Zero

1

8.854 x 10 − 12 C 2 N − 1 m − 2

8.854 x 1012 C2 N-1 m-2

Question 10

gets compressed

gets expanded

oscillates

remains unchanged

Solution

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V²/2em

W _A < W_B < W_C

W_A = W_B= W_C

W_A > W_B > W_C

W_A = W_B < W_C

q /∈₀

q /6∈₀

q/8∈₀

q/ 32∈₀

C₂ V/ C₁ + C₂

C₁V/ C₁ + C₂

1 + C₂/ C₁

(1 - C₂/ C₁)V

8.854 x 10^{− 12}C² N^{− 1} m^{− 2}

8.854 x 10¹²C² N^-1 m^-2