Physics Test

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Physics Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is R. What will be new radius, when energy of particle is doubled?

A
3R

B
2R

C
R√3

D
R√2

Solution

In this case,

$r=\frac{\sqrt{2 m k}}{q B}=r \propto \sqrt{K}$
$\Rightarrow \frac{R}{R_{2}}=\sqrt{\frac{K}{2 K}} \Rightarrow R_{2}=R \sqrt{2}$
Question 2
1 / -0

An oil drop having charge 2e is placed stationary between two parallel horizontal plates 2 cm apart when a potential difference of 12000 volts is applied between them. What will be the radius of the drop, if the density of oil is 900 kg/m3?

A
1.1 x 10^-6 m

B
1.4 x 10^-6 m

C
1.7 x 10^-6 m

D
2 x 10^-6 m

Solution

In equilibrium, $Q E=m g$ $\Rightarrow \mathrm{Q}_{\text {rac }} \mathrm{V} \mathrm{d}=\mathrm{m} \mathrm{g}=\left(\operatorname{rac} 43 \pi \mathrm{r}^{3} \text { hoight }\right) \mathrm{g}$
\[
\begin{array}{l}
\Rightarrow 2 \times 1.6 \times 10^{-19} \times \frac{12000}{2 \times 10^{-2}}=\frac{4}{3} \pi \mathrm{r}^{3} \times 900 \times 10 \\
\Rightarrow \mathrm{r}=1.7 \times 10^{-6} \mathrm{m}
\end{array}
\]
Question 3
1 / -0

A source of e.m.f. E = 15 V and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as i = 1.2 t + 3. Then, find the total charge that will flow in first 5 second.

A
40 C

B
30 C

C
20 C

D
10 C

Solution

Current, $i=\frac{d Q}{d t} \Rightarrow d Q=$ idt
$\Rightarrow Q=\int_{t_{1}}^{t_{2}} i d t=\int_{0}^{5}(1.2 t+3) d t$
$=\left[\frac{1.2 t^{2}}{2}+3 t\right]_{0}^{5}=30 \mathrm{c}$
Question 4
1 / -0

When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, then what will be the effect on its resistance?

A
16 times

B
8 times

C
4 times

D
12 times

Solution

Because, in stretching of wire R∝1/r⁴
Question 5
1 / -0

If a proton, deuteron and α- particle on being accelerated by the same potential difference enters perpendicular to the magnetic field, then find the ratio of their kinetic energies.

A
1:1:2

B
1:2:1

C
1:2:2

D
2:2:1

Solution

The kinetic energy in magnetic field remains constant and is,

$K=q V \Rightarrow K \propto q \quad(V=c o n$ stant
$\therefore K_{p}: K_{d}: K_{a}=q_{p} ; q_{d} ; q_{a}=1: 1: 2$
Question 6
1 / -0

2 rods of same material and length have their electric resistance in ratio 1:2. Which of the following will be the correct statement, when both rods are dipped in water?

A
Loss of weight will be in the ratio 1:2

B
A has more loss of weight

C
B has more loss of weight

D
Both have same loss of weight

Solution

$\mathrm{A} \mathrm{S} \mathrm{R}=\rho \frac{1}{\mathrm{A}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{A}_{2}}{\mathrm{A}_{2}} \quad(\rho, \mathrm{L} \text { constant })$
$\Rightarrow \frac{A_{1}}{A_{2}}=\frac{R_{2}}{R_{1}}=2$
Now, when a body dipped in water, loss of weight $=\mathrm{Vo}_{\mathrm{L}} \mathrm{g}=\mathrm{AL} \sigma_{\mathrm{L}} \mathrm{g}$
$\therefore \frac{(\text { Loss of weight })_{1}}{ \text { (Loss of weight } )_{2}}=\frac{A_{1}}{A_{2}}=2:$ so A has more loss of weight.
Question 7
1 / -0

The figure shows, three charges Q, (+q) and (+q) that are situated at the vertices of an equilateral triangle of side 'l'. If the net electrostatic energy of the system is zero, then evaluate Q.

A
zero

B
(+q)

C
(-q)

D
(-q/2)

Solution
Question 8
1 / -0

A particle has 'a' mass 400 times than that of the electron and charge is double than that of a electron. It is accelerated by 5V of potential difference. Initially the particle was at rest, then what will be its final kinetic energy?

A
200 eV

B
100 eV

C
10 eV

D
5 eV

Solution

Here, $∆ E = 2 e \times 5 V = 10 e V$

$\Rightarrow$ Final kinetic energy = 10 eV
Question 9
1 / -0

A piece of wire of resistance 4 ohms is bent through 180^o at its mid-point and two halves are twisted together, then what is its resistance?

A
1 ohm

B
2 ohm

C
5 ohm

D
6 ohm

Solution

Two halves each of resistance 2Ω are in parallel in twisted wire, therefore equivalent resistance will be 2/2 = 1Ω
Question 10
1 / -0

In a wire of circular cross-section with radius r, free electrons travel with a drift velocity V when a current I flows through the wire. When the drift velocity is 2V, then how much is the current in another wire of half the radius and of the same material?

A
1/4

B
1/2

C
1

D
21

Solution

$\mathrm{From}, \mathrm{v}_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}} \Rightarrow \mathrm{i} \propto \mathrm{v}_{\mathrm{d}} \mathrm{A}$ i.e. $\mathrm{i} \propto \mathrm{v}_{\mathrm{d}} \mathrm{r}^{2}$

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Physics Test - 29

Result

Physics Test - 29

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SHARING IS CARING

Question 1

3R

2R

R√3

R√2

Solution

Question 2

1.1 x 10-6 m

1.4 x 10-6 m

1.7 x 10-6 m

2 x 10-6 m

Solution

Question 3

40 C

30 C

20 C

10 C

Solution

Question 4

16 times

8 times

4 times

12 times

Solution

Question 5

1:1:2

1:2:1

1:2:2

2:2:1

Solution

Question 6

Loss of weight will be in the ratio 1:2

A has more loss of weight

B has more loss of weight

Both have same loss of weight

Solution

Question 7

zero

(+q)

(-q)

(-q/2)

Solution

Question 8

200 eV

100 eV

10 eV

5 eV

Solution

Question 9

1 ohm

2 ohm

5 ohm

6 ohm

Solution

Question 10

1/4

1/2

1

21

Solution

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1.1 x 10^-6 m

1.4 x 10^-6 m

1.7 x 10^-6 m

2 x 10^-6 m