Physics Test

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Physics Test - 13

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Question 1
1 / -0

Two small charged spheres A and B have charges 10 µC and 40 µC, respectively, and are held at a separation of 90 cm from each other. At what distance from A would the electric intensity be zero?

A
22.5 cm

B
18 cm

C
36 cm

D
30 cm

Solution

$E_{A}=\frac{k q_{A}}{x^{2}}$ rightward
And $E_{B}=\frac{k q_{B}}{(l-x)^{2}}$ leftward
since, there is no field at $\mathrm{P}$.
$E=E_{A}-E_{B}=0$
$\Longrightarrow k q\left(\frac{q_{A}}{x^{2}}-\frac{q_{B}}{(l-x)^{2}}\right)=0$
$\Longrightarrow\left(\frac{l-x}{x}\right)^{2}=\frac{q_{B}}{q_{A}}$
$\Longrightarrow \frac{l-x}{x}=\sqrt{\frac{40}{10}}=2$
$\Longrightarrow l-x=2 x$
$\Longrightarrow x=\frac{l}{3}=\frac{90 \mathrm{cm}}{3}$
$\Longrightarrow x=30cm$
Question 2
1 / -0

If the distance between two point charges is made 3 times,the new force on either charge will be_______.

A
1/9th of the original value

B
1/3rd of the original value

C
1/5th of the original value

D
1/7th of the original value

Solution

Initial case when distance between charges is $r$ the force can be determined as, $\vec{F}_{}=\frac{K q_{1} q_{2}}{r^{2}}$
But when the radius between the charges increases that is when $r \rightarrow 3 r$, New distance b/w charges, $r'=3 r,$ then force can be defined as,
$\vec{F'}=\frac{K q_{1} q_{2}}{\left(r^{1}\right)^{2}}=\frac{K q_{1} q_{2}}{(3 r)^{2}}=\frac{K q_{1} q_{2}}{9 r^{2}}$
$\vec{F'}=\frac{\vec{F}}{9}$
As the distance increases by 3 times the force reduces by 9 times.
Question 3
1 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.
three charges -q, Q, -q are placed at a equal distance on a staright line. if the total potential energy of the system is zero, then what is the ratio of Q:q ?

A
1:4

B
1:2

C
2:1

D
4:1

Solution

Current I will be independent of the resistance $\mathrm{R}_{6}$ if resistance $\mathrm{R}_{6}$ is ineffective. This will be possible only when resistances $R_{1}, R_{2}, R_{3}$ and $R_{4}$ form a balanced Wheatstone bridge, i.e. $\frac{R_{1}}{R_{2}}=\frac{R_{3}}{R_{4}}$ or $R_{1} R_{4}=R_{2} R_{3}$
Question 4
1 / -0

A parallel plate capacitor has a voltage of 5 V between the two plates separated by 4 microns (1 micron = 10^-6 m). What is the electric field between the capacitor plates?

A
1.02 × 10⁵ V/m

B
1.25 × 10⁶V/m

C
3.4 × 10⁴V/m

D
4.3 × 10⁴V/m

Solution

Power of each bulb $=60 \mathrm{W}$

When three bulbs are connected in series, power drawn will be $\frac{1}{P^{\prime}}=\frac{1}{60}+\frac{1}{60}+\frac{1}{60}$ or $\frac{3}{60}$

Or $\frac{1}{20}=\frac{1}{P^{\prime}}$ or $P^{\prime}=20 W$
Question 5
1 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.

Two-point charges 2x10-6C & 4x10-6C are 12 cm apart. what is the amount of work done to bring the charges 3 cm closer?

A
1.8 J

B
1.6 J

C
2.4 J

D
0.6 J

Solution

Net resistance of the circuit = r₁ + r₂ + R
Net emf in series = E + E = 2E
Therefore, from Ohm's law, current in the circuit
$\begin{aligned} i &=\frac{\text { net emf }}{\text { net resistance }} \\ \Rightarrow i &=\frac{2 E}{r_{1}+r_{2}+R} \end{aligned}$
It is given that, as circuit is closed, potential difference across the first cell is zero. That is, $V=E-i r_{1}=0$
$\Rightarrow \quad i=\frac{E}{r_{1}}$
Equation Equations (i) and (ii), we get
\[
\begin{aligned}
& \frac{E}{r_{1}}=\frac{2 E}{r_{1}+r_{2}+R} \\
\Rightarrow & 2 r_{1}=r_{1}+r_{2}+R \\
R=& \text { external resistance }=r_{1}-r_{2}
\end{aligned}
\]
Question 6
1 / -0

Two charges of equal magnitudes at a distance 'r' exert a force 'F' on each other. If the charges are halved and the distance between them is doubled, then the new force acting on each charge is

A
F/8

B
F/4

C
4F

D
F/16

Solution
Question 7
1 / -0

Two wires A and B, made of same material and having their lengths in the ratio 6 : 1 are connected in series. The potential differences across the wires are 3 V and 2V, respectively. If r_A and r_B are the radii of A and B respectively, then $\frac{\mathrm{r}_{\mathrm{P}}}{\mathrm{r}_{\mathrm{A}}}$ is

A
^{$\frac{1}{4}$}

B
^{$\frac{1}{2}$}

C
1

D
2

Solution

shunt resistance $R s=\frac{R g^{*} 1 g}{1-l g} ; \Rightarrow R s=\frac{13 \times 100}{750-100}=\frac{1300}{650}=2 \Omega$
Question 8
1 / -0

The orbital angular momentum of a satellite revolving at a distance 'r' from the centre of Earth is 'L'. If the distance is increased to '16 r', what will be the orbital angular momentum?

A
12 L

B
4 L

C
8 L

D
L/4

Solution

Displacement $\vec{s}=10 \hat{j} \mathrm{m} \text { (Along the } y \text { -axis only })$

Force $\vec{F}=(2 i+15 \hat{j}+6 \hat{k}) N$

Work done

$W=\vec{F} \cdot \vec{s}=(2 \hat{i}+15 \hat{j}+6 \hat{k}) \cdot(10 \hat{j})=150 J$
Question 9
1 / -0

The time period of a satellite of Earth is 5 hours. If the separation between Earth and the satellite is increased to 4 times the previous value, the new time period will become

A
10 hours

B
80 hours

C
40 hours

D
20 hours

Solution

Power $=\frac{\text { energy }}{\text { time }}$

$=\frac{m g h}{t}=\frac{6000 \times 10 \times 100}{60} \mathrm{W}$

Power $=100 \mathrm{kW}$
Question 10
1 / -0

In order to shift a body of mass 'm' from a circular orbit of radius '3R' to a higher orbit of radius '5R' around Earth, where 'R' is radius of Earth; the work done is:

A
$\frac{3 G M m}{5 R}$

B
$\frac{G M m}{2 R}$

C
$\frac{2 G M m}{15 R}$

D
$\frac{G M m}{5 R}$

Solution

From the principle of conservation of energy, we have
$\frac{1}{2} m v^{2}+m g h=0+m g R$
which gives $v=\sqrt{2 g(R-H)}$
Here, $H=h \mathrm{So}, \mathrm{v}=(2 \mathrm{g}(\mathrm{R}-\mathrm{h}))^{1 / 2}$

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Physics Test - 13

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Physics Test - 13

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SHARING IS CARING

Question 1

22.5 cm

18 cm

36 cm

30 cm

Solution

Question 2

1/9th of the original value

1/3rd of the original value

1/5th of the original value

1/7th of the original value

Solution

Question 3

1:4

1:2

2:1

4:1

Solution

Question 4

1.02 × 105 V/m

1.25 × 106 V/m

3.4 × 104 V/m

4.3 × 104 V/m

Solution

Question 5

1.8 J

1.6 J

2.4 J

0.6 J

Solution

Question 6

F/8

F/4

4F

F/16

Solution

Question 7

14

12

1

2

Solution

Question 8

12 L

4 L

8 L

L/4

Solution

Question 9

10 hours

80 hours

40 hours

20 hours

Solution

Question 10

\(\frac{3 G M m}{5 R}\)

\(\frac{G M m}{2 R}\)

\(\frac{2 G M m}{15 R}\)

\(\frac{G M m}{5 R}\)

Solution

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1.02 × 10⁵ V/m

1.25 × 10⁶V/m

3.4 × 10⁴V/m

4.3 × 10⁴V/m

^{$\frac{1}{4}$}

^{$\frac{1}{2}$}