Physics Test-8

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Physics Test-8

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Question 1
4 / -1

Calculate the force between an alpha particle and a proton separated by $5.12 \times 10^{-15} m$.

A
$11.5 N$

B
$15.5 N$

C
$17.5 N$

D
$20.5 N$

Solution

Given that:

Separation distance, $r=5.12 \times 10^{-15} m$

We know that:

Charge on an electron, $e=1.6 \times 10^{-19}$

and, $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9}$

Charge on an alpha particle is $2 e$.

$\therefore$ Using Coulomb's law,

$F=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$=\frac{1}{4 \pi \epsilon_0} \frac{2 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(5.12 \times 10^{-15}\right)^{2}}$

$=9 \times 10^{9} \times 0.195 \times 10^{-8}$

$=17.5 N$
Question 2
4 / -1

Which of the following is a bad conductor of electricity?

A
Distilled water

B
Silver nitrate

C
Sulphuric acid

D
Copper sulphate

Solution

Distilled wateris a bad conductor of electricity.

In distilled water, there are no impurities, there are no ions, there are only neutral (no charge) water molecules and these neutral molecules don't have charge, so distilled water does not conduct electricity.
Question 3
4 / -1

A free neutron decays into a proton, an electron and

A
a neutrino

B
an antineutrino

C
an $\alpha$ - particle

D
a $\beta$ - particle

Solution

A neutron is converted into a proton in beta minus, increasing the atomic number of the atom. The neutron is a neutral particle, while the proton is a positive particle. In order to preserve charge conservation, the nucleus also releases an electron and an antineutrino. The antimatter equivalent of neutrino is antineutrino. Both of these particles are massless and neutral. They have a very weak interaction with matter and can travel through the entire earth without being disturbed.
Question 4
4 / -1

The speed of electromagnetic waves in vacuum depends upon the source of radiation. It:

A
Increases as we move from $\gamma$-rays to radio waves

B
Decreases as we move from $\gamma$-rays to radio waves

C
Is same for all of them

D
None of these

Solution

The speed of electromagnetic waves in vacuum depends upon the source of radiation. It is same for all of them.

We know that speed of electromagnetic wave in vacuum is given by,

Speed of light = Frequency × Wavelength

$c=\nu \times \lambda$

$=\frac{1}{\mu_0 \varepsilon_0}$

= Constant

As we go from gamma rays to radio waves frequency decreases and wavelength increases thereby maintaining the product constant.
Question 5
4 / -1

The temperature of a place on one sunny day is 113 in Fahrenheit scale. The Kelvin scale reading of this temperature will be

A
318 K

B
45 K

C
62.8 K

D
335.8 K

Solution

It is given that the temperature of a place on one sunny day is 113 in Fahrenheit scale.

Relation between Fahrenheit and Kelvin is-

$\frac{F - 32^{\circ}}{9} = \frac{K - 273}{5}$

Therefore, according to the question,

$\begin{array}{l} \Rightarrow \frac{113 - 32^{\circ}}{9} = \frac{K - 273}{5} \\ \Rightarrow 9 = \frac{K - 273}{5} \\ \Rightarrow K = 318 K \end{array}$
Question 6
4 / -1

Specific resistance of all metals is most affected by:

A
Temperature

B
Pressure

C
Degree of illumination

D
Applied magnetic field

Solution

Specific resistance of metal at temperature T,$\rho_{\mathrm{T}}=\rho_0(1+\alpha \Delta \mathrm{T})$

Specific resistance of metal depends on the temperature of the metal and it increases with increase in temperature. Also specific resistance of metals does not vary with volume of substance, pressure and applied magnetic field.
Question 7
4 / -1

A uniform rope of length $10 \mathrm{~m}$ and mass $3 \mathrm{~kg}$ hangs vertically from a rigid support. A block of mass $1 \mathrm{~kg}$ is attached to the free end of the rope. A transverse pulse of wavelength $0.05 \mathrm{~m}$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is:

A
0.10 m

B
0.12 m

C
0.16 m

D
0.18 m

Solution

Since the rope has a mass, the tension along its length is a variable.

At the top end, its tension $T_{\text{Top}}=(3+1) g \mathrm{~N}$

$= 4 \mathrm{~g} \mathrm{~N}$

At the bottom end, its tension $T_{\text{Bottom}}=1 g \mathrm{~N}$

Now, the speed of the transverse wave on the string,

$c=\sqrt{\frac{T}{\mu}}$

$\Rightarrow v \lambda=\sqrt{\frac{T}{\mu}} \quad(\because c=v \lambda)$

$\frac{\sqrt{T_{\text {Top }}}}{\lambda_{\text {Top }}}=\frac{\sqrt{T_{\text {Bottom }}}}{\lambda_{\text {Bottom }}}$

$\Rightarrow\frac{\sqrt{4 g}}{\lambda_{\text {Top }}}=\frac{\sqrt{1 g}}{0.05}$

$\Rightarrow \lambda_{\text {Top }}=0.05 \times 2 \mathrm{~m}$

$\Rightarrow \lambda_{\text {Top }} =0.10 \mathrm{~m}$
Question 8
4 / -1

The value of acceleration due to gravity at Earth's surface is $9.8 \mathrm{~ms}^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9 \mathrm{~ms}^{-2}$, is close to: (Radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ )

A
$1.6 \times 10^{6} \mathrm{~m}$

B
$9.0 \times 10^{6} \mathrm{~m}$

C
$6.4 \times 10^{6} \mathrm{~m}$

D
$2.6 \times 10^{6} \mathrm{~m}$

Solution

Given:

Acceleration due to gravity at Earth's surface is $g_{0}=9.8 \mathrm{~ms}^{-2}$.

Acceleration due to gravity decreases to $g=4.9 \mathrm{~ms}^{-2}$.

We know that radius of earth (R) $=6.4 \times 10^{6}$ m

Let height is h

$g=\frac{g_{0}}{\left(1+\frac{h}{R}\right)^{2}}$

Put all the given values in above formula:

$4.9=\frac{9.8}{\left(1+\frac{h}{6.4 \times 10^{6}}\right)^{2}}$

$\Rightarrow 1+\frac{h}{6.4 \times 10^{6}}=\sqrt{2}$

$\Rightarrow h=(\sqrt{2}-1) 6.4 \times 10^{6} $

$\Rightarrow h=(1.414-1) 6.4 \times 10^{6} $

$\Rightarrow h=2.6 \times 10^{5} \mathrm{~m}$
Question 9
4 / -1

Which of the following setups can be used to verify Ohm's law?

A

B

C

D

Solution

In a circuit ammeter is always connected in series and voltmeter is always connected in parallel. So the option A is the correct setups for verifying Ohm's law.
Question 10
4 / -1

A tank, filled partially with a liquid, is subjected to uniform horizonal acceleration. Which of the following is true about the surface of the liquid in the tank?

A
The surface of liquid remains horizontal

B
The liquid surface falls down only at the center of the tank

C
The liquid surface falls down on the direction of motion and rises up on the back side of the tank

D
None of these
Solution
"The liquid surface falls down on the direction of motion and rises up on the back side of the tank" is true about the surface of the liquid in the tank.
- As we know that if the liquid is accelerated horizontally, then the two points in the same horizontal line can’t have equal pressure.
- Therefore, the free surface of the liquid is inclined to the horizontal with larger depth at the rear end as the tanker is moving in the forward direction with uniform acceleration.