Physics Test 262

Result

Physics Test 262

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Weekly Quiz Competition

Question 1
4 / -1

E, m, p and G denote energy, mass, angular momentum and gravitational constant. Then has the dimensions of

A
Length

B
Mass

C
Angle

D
Time

Solution
Question 2
4 / -1

A circular conducting loop of radius R carries a current I. Another straight infinite conductor carrying current I passes through the diameter of this loop as shown in the figure. The magnitude of force exerted by the straight conductor on the loop is :-

A

B

C

D

Solution
Question 3
4 / -1

Find the intensity of electromagnetic wave if the electric field in electromagnetic wave is

A
13.24 watt/m²

B
16.2 watt/m²

C
25.3 watt/m²

D
32.3 watt/m²

Solution
Question 4
4 / -1

The displacement of a particle is represented by the equation y= 40cos³ωt. The motion is

A
Non Periodic

B
Simple harmonic with period 5π/ω

C
Periodic but not Simple harmonic

D
Simple harmonic with period 2π/ω

Solution
Question 5
4 / -1

In the circuit shown in figure find the current in branch BD of the circuit :

A
5 A

B
0.5 A

C
11/3 A

D
None of these

Solution
Question 6
4 / -1

A

B

C

D

Solution
Question 7
4 / -1

One million small identical drops of water, all charged to the same potential, are combined to form a single large drop. If E is the sum of the electrostatic energy of each small drop, the combined energy of the large drop is

A
10^–4 E

B
E

C
10⁶ E

D
10⁴ E

Solution
Question 8
4 / -1

An unknown particle originally at rest emits 5 alpha particles with speed 11385 km/h. Find the recoil speed of the unknown daughter nucleus.

A
35 m/s

B
50 m/s

C
134 m/s

D
100 m/s

Solution
Question 9
4 / -1

The number of AM broadcast stations that can be accommodated at a 150 kHz band width, if the highest frequency modulating carrier is 5 kHz is

A
15

B
18

C
21

D
24

Solution

Total bandwidth = 150 kHz

f_a(max) = 5 kHz

Any station being modulated by a 5 kHz will produce an upper-side frequency 5 kHz above its carrier and a lower-side frequency 5 kHz below its carrier.

Thus one station needs a bandwidth of 10 kHz.

Number of stations accommodated = Total BW /BW per station

=(150×10³)/(10×10³)=15
Question 10
4 / -1

A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is 36^∘C. Then the difference of temperature at the two surfaces of A will be

A
6^∘C

B
12^∘C

C
18^∘C

D
24^∘C

Solution