Physics Test

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Physics Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

In a transformer 220 A.C. voltage is increased to 2200 volts. If the number of turns in the secondary are 2000, then the number of turns in the primary will be

A
200

B
100

C
50

D
20

Solution

Number of turns in primary coil ( Np ) = ?

Number of turns in Secondary coil ( Ns ) = 2000

Voltage in primary coil ( Ep ) = 220 V

Voltage in secondary coil ( Es ) = 2200 V .

Now, using formula: Ep/Es = Np/Ns

Substituting values, we get,

220/2200 = Np/2000

=> Np = 200 turns .
Question 2
1 / -0

In a photo electric experiment the maximum velocity of photo electons emitted

A
Depends on intensity of incident radiation

B
Does not depend on the cathode material

C
Depends on frequency of incident radiation

D
Does not depend on wave length of incident raidation
Question 3
1 / -0

Two wires of same metal have the same length but their cross sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination is

A
5/2 Ω

B
40/3 Ω

C
40 Ω

D
100 Ω

Solution
Question 4
1 / -0

If elements with principal quantum number n > 4 were not allowed in nature, then, the number of possible elements would be

A
32

B
60

C
18

D
4

Solution

Correct option is B)

Hint:

Add all the elements from n=1 to n=4.

Step 1: Write the number of elements for each shell.

The number of elements for n=1 are 2.

The number of elements for n=2 are 8.

The number of elements for n=3 are 18.

The number of elements for n=4 are 32.

Step 2: adding all the elements,

from n=1 to n=4.

Total number =2+8+18+32=60

Hence option B correct
Question 5
1 / -0

If a cell of constant electromotive force produces the same amount of heat during the same time in two independent resistances R₁ and R₂, when connected separately one after the other, across the cell, then the internal resistance of the cell is

A
(R₁ + R₂)

B
(R₁ - R₂)/2

C
√R₁ R₂/2

D
√R₁ R₂

Solution

Answer :- d

Solution :- Here current through the resistance R1 is I1 = ε/(r+R1)

and current through the resistance R2 is I2 = ε/(r+R2).

Also, the heat produced due to passage of current through both the resistances is same.

Hence, Q = (I1)^2R1 = (I2)^2R2

Using the values of I1 and I2 in above equation we get (ε/(r+R1))^2 R1 = (ε/(r+R2))^2 R2

where, ε is the emf and r is the internal resistance of battery.

ε^2 R1 / (r+R1)^2 = ε^2 R1 / (r+R1)^2

r^2(R1−R2) = R1R2((R1−R2)

r^2= R1R2

∴ r = (R1R2)^1/2
Question 6
1 / -0

A wire of radius r has resistance R. If it is stretched to a radius of 3r/4, its resistance becomes

A
9R/16

B
16R/9

C
81R/256

D
256R/81
Question 7
1 / -0

Optical fibres uses the phenomenon of

A
total internal reflection

B
refraction

C
dispersion

D
scattering
Question 8
1 / -0

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

A
1/2 ε₀ E² Ad

B
1 2ε₀ E² /Ad

C
ε₀ E² /Ad

D
ε₀ E² Ad

Solution

Energy required = 1 2 CV² = 1 2 ε₀ E²Ad
Question 9
1 / -0

Determine the current drawn from a 12 v supply with internal resistance 0.5Ω by the infinite network shown in the figure . Each resistor has Ω resistance

A
2.71

B
3.30

C
3.72

D
5.24

Solution

Let the equivalent resistance of the network be x×1

R_s=1+1+1+1

x = xΩ

The effective resistance x and 1Ω are in parallel
Question 10
1 / -0

A cell whose e.m.f. is 2V and internal resistance is 0.1Ω, is connected with a resistance of 3.9Ω. The voltage across the cell terminal will be

A
0.50 V

B
1.90 V

C
1.95 V

D
2.00 V

Solution

Answer :- c

Solution :- EMF of the cell(E) =2 V

internal resistance(r)= 0.1 V

External resistance(R)=3.9 V

V=IR

=(E÷(R+r))×R

=(2÷(3.9+0.1))×3.9

=1.95 V

Therefore potential difference across the terminals of the cell = 1.95V

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Physics Test - 1

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Physics Test - 1

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SHARING IS CARING

Question 1

200

100

50

20

Solution

Question 2

Depends on intensity of incident radiation

Does not depend on the cathode material

Depends on frequency of incident radiation

Does not depend on wave length of incident raidation

Question 3

5/2 Ω

40/3 Ω

40 Ω

100 Ω

Solution

Question 4

32

60

18

4

Solution

Question 5

(R₁ + R₂)

(R₁ - R₂)/2

√R₁ R₂/2

√R₁ R₂

Solution

Question 6

9R/16

16R/9

81R/256

256R/81

Question 7

total internal reflection

refraction

dispersion

scattering

Question 8

1/2 ε0 E2 Ad

1 2ε0 E2 /Ad

ε0 E2 /Ad

ε0 E2 Ad

Solution

Question 9

2.71

3.30

3.72

5.24

Solution

Question 10

0.50 V

1.90 V

1.95 V

2.00 V

Solution

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1/2 ε₀ E² Ad

1 2ε₀ E² /Ad

ε₀ E² /Ad

ε₀ E² Ad