Physics Test

Result

Physics Test - 34

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
5 / -1

Three similar light bulbs are connected to a constant voltage d.c. supply as shown in Fig. Each bulb operates at normal brightness and the ammeter (of negligible resistance) registers a steady current.

The filament of one of the bulbs breaks. What happens to the ammeter reading and to the brightness of the remaining bulbs?

Ammeter reading _______ & Bulb brightness _______

A
increases, increases

B
increases, unchanged

C
unchanged, unchanged

D
decreases, unchanged

Solution

Suppose V is the voltage of the supply and R is the resistance of each bulb.

Now, R_p = R/3 and current in ammeter, I = V/R_p = 3 V/R, provided all three bulbs are working properly.

If one bulb has broken down, then R_p = R/2 and I = 2V/R

Therefore, current decreases and since current through each bulb is V/R the same as before, brightness of bulbs is not affected.
Question 2
5 / -1

In a meter bridge experiment a balance point is obtained at a distance of 60 cm from the left end when unknown resistance R is in a left gap and 8 ohms resistor is connected in the right gap. When the position of R and 8 ohm resistor is interchanged the balance point will be at distance of

A
40 cm

B
30 cm

C
50 cm

D
60 cm

Solution

60/40 = l/100-l, = R/8

so R=12

now changed position 8 is proportional to l and 12 to (100-l)

8/12 = l/100-l

l = 40ohm
Question 3
5 / -1

V-I graph of which material shows the straight line

A
Silicon

B
Silver

C
Germanium

D
Gallium

Solution

Materials which obeys the ohm’s law show straight line in the V-I graph. Since silver is the only ohmic material in given options, it shows straight line curve.
Question 4
5 / -1

The following fig. shows I-V graph for a given metallic wire at two temperatures T₁and T₂.Then,

A
Temperature T₂ is less

B
Temperature T₂ is more

C
T₁ is same as T₂

D
None of the above

Solution

The slope of the graph = I / V

In case of T₂, I/V is less. Hence, V/I (=R) is more.

Value of resistance increases with increasing temperature.

Hence, T₂ >T
Question 5
5 / -1

In two current carrying conductors parallel currents________, anti parallel currents_________ .

A
attract , attract

B
attract , repel

C
repel , attract

D
repel , repel

Solution

parallel currents attract from the below figure.

and similarly, in antiparallel currents repel. parallel currents attract from the below figure.

and similarly, in antiparallel currents repel.
Question 6
5 / -1

No force is exerted by a magnetic field on a stationary –

A
current carrying conductor

B
current loop

C
electric dipole

D
magnetic dipole
Solution
No force is exerted by a magnetic field on a stationary electric dipole because:
- The electric dipole consists of two equal and opposite charges.
- When stationary, the dipole does not experience any motion through the magnetic field.
- As a result, there is no induced current or change in magnetic flux.
- The magnetic field only affects moving charges or current-carrying conductors.
Therefore, the correct understanding is that a magnetic field does not exert a force on a stationary electric dipole.
Question 7
5 / -1

Along an infinitely long conductor carrying a current of 8 A we keep another conductor of length 5 m carrying a current of 3 A. Both the conductors are 10 cm apart. Find the force on small conductor.

A
2.4 X 10^-4 N

B
9.6 X 10^-4 N

C
2.6 X 10^-6 N

D
9.6 X 10^-5 N

Solution

I₁=3A

I₂=8A

l=5m

r=0.1m

F=?

Force per unit length on the small conductor is given by:

f= μ_o2I₁I₂/4πr

Total force on the length of small conductor is

F=fl

F= μ_o2I₁I₂ l/4πr

F=4πx10^-7x2x3x8x5/4π x 0.1

F=2.4x10^-4N
Question 8
5 / -1

We use _________ to find the direction of force on a current-carrying conductor placed in a magnetic field.

A
Lenz’s Law

B
Right Hand Rule

C

D
Ampere’s Circuital Law

Solution

According to NCERT, the force on a current-carrying conductor in a magnetic field is given by

The direction is found using the right-hand rule for vector (cross) product: point your fingers along l (direction of current) and curl them towards B; your thumb gives the direction of force.
Question 9
5 / -1

What is the value of power factor in RLC circuit

A

B

C

D
Noe of the above

Solution
Question 10
5 / -1

In a series LCR what will be phase difference between voltage drop across inductor and capacitor

A
0

B
90

C
180

D
45

Solution

Let’s keep this simple and to the point. We know that:

1.in a series circuit the same current flows through each component

2.the voltage across an ideal inductor L is 90˚ ahead of an AC current through it

3.the voltage across an ideal capacitor C is 90˚ behind an AC current through it

So putting these facts together we can conclude that given an AC series current the voltages across any L and C must have a phase difference of 180˚