Physics Test

Result

Physics Test - 21

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
1 / -0

A proton of energy 200 MeV enters the magnetic field of 5 T. If direction of field is from south to north and motion is upward, then how much force will be acting on it?

A
1.6 x 10^-6 N

B
3.2 x 10^-8 N

C
1.6 x 10^-10 N

D
0

Solution

Force, \(F=q v B\) also Kinetic energy, \(\kappa=\frac{1}{2} m v^{2}\)
\(\Rightarrow v=\sqrt{\frac{2 K}{m}}\)
\(\therefore F=q \sqrt{\frac{2 K}{m}} B\)
\(\Rightarrow F=1.6 \times 10^{-19} \sqrt{\frac{2 \times 200 \times 10^{6} \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}} \times 5\)
\(\Rightarrow F=1.6 \times 10^{-10} \mathrm{N}\)
Question 2
1 / -0

An α particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their velocities, then what is the ratio of the radii of their circular path?

A
1:2

B
2:1

C
1:4

D
4:1

Solution

Here, \(r=\frac{m v}{q B}\)
\(\Rightarrow \frac{r_{a}}{r_{p}}=\frac{m_{a}}{m_{p}} \times \frac{q_{p}}{q_{a}}\)
\(\Rightarrow \frac{r_{a}}{r_{p}}=\frac{4}{1} \times \frac{1}{2}=\frac{2}{1}\)
Question 3
1 / -0

A wire of length 5m and radius 1mm has a resistance of 1 ohm. What length of wire of the same material at same temperature and of radius 2mm will also have a resistance of 1 ohm?

A
20 m

B
10 m

C
2.5 m

D
1.25 m

Solution

since, \(R \propto \frac{1}{r^{2}}\)
\(\therefore \frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}\)
\(\Rightarrow \frac{1}{1}=\frac{5}{l_{2}} \times\left(\frac{2}{1}\right)^{2} \Rightarrow l_{2}=20 \mathrm{m}\)
Question 4
1 / -0

A very long straight wire carries a current I. When a charge +Q at point P has velocity at the instant as shown in fig., then what is the force on the charge?

A
Along OY

B
Opposite to OY

C
Along OX

D
Opposite to OX

Solution

The magnetic field produced by wire at the location of charge is perpendicular to the paper inwards. Thus, by applying Fleming's left hand rule, force is directed along OY.
Question 5
1 / -0

A cube of side 'b' has a charge 'q' at each of its vertices. What will be electric field due to this charge distribution at the centre of this cube?

A
Zero

B
32q/b²

C
q/2b²

D
q/b²

Solution

Because of symmetric charge distribution.
Question 6
1 / -0

What is the maximum kinetic energy of the positive ion in the cyclotron?

A

\(\frac{\mathrm{qBr}_{0}}{2 \mathrm{m}^{2}}\)

B

\(\frac{q^{2} B^{2} r_{0}^{2}}{2 m}\)

C

\(\frac{\mathrm{qB}^{2} \mathrm{r}_{\mathrm{o}}}{2 \mathrm{m}}\)

D

\(\frac{q^{2} \operatorname{Br}_{0}}{2 m}\)

Solution

Maximum kinetic energy,
\(K_{\max }=\frac{1}{2} m v^{2}\)
and \(r_{0}=\frac{m v}{q B} \Rightarrow v=\frac{q B r_{0}}{m}\)\[
\therefore K_{\max }=\frac{1}{2} m\left(\frac{q B r_{0}}{m}\right)^{2}=\frac{q^{2} B^{2} r_{0}^{2}}{2 m}
\]
Question 7
1 / -0

4 wires each of length 2 meter are bent into 4 loops P, Q, R and S and then suspended into uniform magnetic field. Same current is passed in each loop. Which of the following statement is correct?

A
Couple on loop S will be the highest

B
Couple on loop R will be the highest

C
Couple on loop Q will be the highest

D
Couple on loop P will be the highest

Solution

Couple of force on loop S will be maximum since for same perimeter the area of loop will be maximum and magnetic moment of loop =i xA .Therefore, it will also be maximum for loop S.
Question 8
1 / -0

Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths are in the ratio 3:2:1. What will be the ratio of electrical resistances?

A
1:2:3

B
1:4:9

C
9:4:1

D
27:6:1

Solution

\(\mathrm{As}, \mathrm{R} \propto \frac{\mathrm{I}^{2}}{\mathrm{m}} \Rightarrow \mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3}=\frac{\mathrm{I}_{1}^{2}}{\mathrm{m}_{2}}: \frac{\mathrm{I}_{2}^{2}}{\mathrm{m}_{2}}: \frac{\mathrm{I}_{3}^{2}}{\mathrm{m}_{3}}\)
\(\therefore \mathrm{R}_{1}: \mathrm{R}_{2}: \mathrm{R}_{3}=\frac{9}{1}: \frac{4}{2}: \frac{1}{3}=27: 6: 1\)
Question 9
1 / -0

2 wires are made up of two different materials whose specific resistance are in the ratio 2:3, length 3:4 and area 4:5. Find the ratio of their resistances.

A
1:2

B
5:8

C
6:8

D
6:5

Solution

Resistance, \(R=\rho \frac{1}{A}\)
\(\Rightarrow \frac{R_{1}}{R_{2}}=\frac{\rho_{1}}{\rho_{2}} \times \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}}=\frac{2}{3} \times \frac{3}{4} \times \frac{5}{4}=\frac{5}{8}\)
Question 10
1 / -0

An electron having charge 'e' and mass 'm' is moving in a uniform electric field E. What will be its acceleration?

A
mE/e

B
eE/m

C
E²e/m

D
e²/m

Solution

Acceleration, a=F/m= eE/m

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 21

Result

Physics Test - 21

Score

-

Rank

-

SHARING IS CARING

Question 1

1.6 x 10-6 N

3.2 x 10-8 N

1.6 x 10-10 N

0

Solution

Question 2

1:2

2:1

1:4

4:1

Solution

Question 3

20 m

10 m

2.5 m

1.25 m

Solution

Question 4

Along OY

Opposite to OY

Along OX

Opposite to OX

Solution

Question 5

Zero

32q/b2

q/2b2

q/b2

Solution

Question 6

\(\frac{\mathrm{qBr}_{0}}{2 \mathrm{m}^{2}}\)

\(\frac{q^{2} B^{2} r_{0}^{2}}{2 m}\)

\(\frac{\mathrm{qB}^{2} \mathrm{r}_{\mathrm{o}}}{2 \mathrm{m}}\)

\(\frac{q^{2} \operatorname{Br}_{0}}{2 m}\)

Solution

Question 7

Couple on loop S will be the highest

Couple on loop R will be the highest

Couple on loop Q will be the highest

Couple on loop P will be the highest

Solution

Question 8

1:2:3

1:4:9

9:4:1

27:6:1

Solution

Question 9

1:2

5:8

6:8

6:5

Solution

Question 10

mE/e

eE/m

E2e/m

e2/m

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

1.6 x 10^-6 N

3.2 x 10^-8 N

1.6 x 10^-10 N

32q/b²

q/2b²

q/b²

E²e/m

e²/m