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Physics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

A metal plate of area \(10^{3} \mathrm{~cm}^{2}\)rests on a layer of oil \(6\) mm thick. A tangential force of \(10^{2} \mathrm{~N}\)is applied on it to move it with a constant velocity of \(6 \mathrm{~cm} \mathrm{~s}^{1}\). The coefficient of viscosity of the liquid is

A
\(0.1\) poise

B
\(0.5\) poise

C
\(0.7\) poise

D
\(0.9\) poise

Solution

\(\eta=\frac{F}{A(\frac {d v} {d y})}\)
\(\therefore \eta=\frac{10^{-2}}{\left(10^{3} \times 10^{-4}\right)\left(\frac{6 \times 10^{-2}}{6 \times 10^{-3}}\right)}\)

\(=\frac{10^{-2} \times 6 \times 10^{-3}}{10^{-1} \times 6 \times 10^{-2}}\)
\(\eta=10^{-2} N sm ^{-2}=0.1\) poise
Question 2
1 / -0

In Young's double slit experiment the \(10^{\text {th }}\) maximum, of wavelength \(\lambda_{1}\) is at a distance of \({y}_{1}\) from the central maximum. When the wavelength of the source is changed to \(\lambda_{2}, 5^{\text {th }}\) maximum is at a distance of \({y}_{2}\) from its central maximum. The ratio \(\frac{{y}_{1}}{{y}_{2}}\)

A

\(\frac{2 \lambda_{1}}{\lambda_{2}}\)

B

\(\frac{2 \lambda_{2}}{\lambda_{1}}\)

C

\(\frac{\lambda_{1}}{2 \lambda_{2}}\)

D

\(\frac{\lambda_{2}}{2 \lambda_{1}}\)

Solution

Position fringe from central maxima:
\({y}_{1}=\frac{{n} \lambda_{1} {D}}{{d}}\)
Given, \({n}=10\)
\(\therefore {y}_{1}=\frac{10 \lambda_{1} {D}}{{d}} . .\) \((i)\)
For second source:
\({y}_{2}=\frac{5 \lambda_{2} {D}}{{d}} . .\) \((ii)\)
\(\therefore \frac{{y}_{1}}{{y}_{2}}=\frac{\frac{10 \lambda_{1} {D}}{{d}}}{\frac{5 \lambda_{2} {D}}{{d}}}\)
\(\Rightarrow \frac{{y}_{1}}{{y}_{2}}=\frac{2 \lambda_{1}}{\lambda_{2}}\)
Question 3
1 / -0

The missing particle in the reaction:

\({ }_{99}^{253} \mathrm{Es}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{101}^{256} \mathrm{Md}+\)

A
Deuteron

B
Proton

C
Neutron

D
\(\beta\) - particle

Solution

Let the particle be \(\mathrm{X}\).

On summing the mass number on left hand side we get \(253+4=257\) units

On right hand side we have 256 units only. So, we should have 1 unit mass number on X element.

On summing the atomic number on left hand side we get \(99+2=101\) units

On right hand side we have 101 units of atomic number. So, we should have 0 unit atomic number on \(\mathrm{X}\) element.

So \({ }_{0}^{1} \mathrm{X}={ }_{0}^{1} \mathrm{n}\) i.e. neutron
Question 4
1 / -0

What is the density of lead under a pressure of \(2.0 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\), if the bulk modulus of lead is \(8.0 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\). Also, the initial density of lead is \(11.4 \mathrm{~g} / \mathrm{cm}^{3}\).

A
\(12.89 \mathrm{~g} / \mathrm{cm}^{3}\)

B
\(14 \mathrm{~g} / \mathrm{cm}^{3}\)

C
\(11.69 \mathrm{~g} / \mathrm{cm}^{3}\)

D
Zero

Solution

Using the definition of Bulk modulus we have

\(\mathrm{K}=\rho \frac{\mathrm{dP}}{\mathrm{d} \rho}\)

or\(\rho_{2}-\rho_{1}=\rho_{1} \frac{\mathrm{dP}}{\mathrm{K}}\)

or\(\rho_{2}=\rho_{1}\left(1+\frac{\mathrm{dP}}{\mathrm{K}}\right)\)

\(=11.4\left(1+\frac{2 \times 10^{8}}{8 \times 10^{9}}\right)\)

\(=11.69 \mathrm{g} / \mathrm{cm}^{3}\)
Question 5
1 / -0

A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are \(\gamma_{1}\) and \(\gamma_{2}\) respectively. If the temperature of both mercury and metal are increased by an amount \(\Delta t\), the fraction of the volume of the metal submerged in mercury changes by the factor.

A
\(\frac{1}{\left(\gamma_{2}-\gamma_{1}\right) \Delta t}\)

B
\(\frac{1}{\left(\gamma_{1}-\gamma_{2}\right) \Delta t}\)

C
\(\left(\gamma_{1}-\gamma_{2}\right) \Delta t\)

D
\(\left(\gamma_{2}-\gamma_{1}\right) \Delta t\)

Solution

Weight of metal piece \(\mathrm{w}_{\mathrm{m}}=\mathrm{mg}\)

When immersed in liquid, let fraction f be submerged in the liquid.

\(\therefore \mathrm{fV}_{0 \mathrm{~m}} \rho_{01} g=\mathrm{mg}\)......(1)

When the metal and the liquid are heated, let f' be the fraction which is submerged in the liquid.

\(\therefore f^{\prime} V_{0 m}\left(1+\gamma_{m} \Delta T\right) \rho_{01}\left(1-\gamma_{1} \Delta T\right) g=m g\)...(2)

Dividing (1) by (2),

\(\frac{f}{f^{\prime}\left(1+\gamma_{m} \Delta T\right)\left(1-\gamma_{1} \Delta T\right)}=1\)

\(\Rightarrow \frac{f^{\prime}}{f}=\left(1-\gamma_{m} \Delta T\right)\left(1+\gamma_{1} \Delta T\right)\)

\(\Rightarrow \frac{f^{\prime}}{f}-1=\left(\gamma_{1}-\gamma_{m}\right) \Delta T\)

\(\frac{f^{\prime}-f}{f}=\left(\gamma_{1}-\gamma_{m}\right) \Delta T=\left(\gamma_{2}-\gamma_{1}\right) \Delta T\)
Question 6
1 / -0

During the melting of solid, its temperature ________.

A
May increase or decrease depending on the nature of solid

B
Decreases

C
Increases

D
Does not change

Solution

During the melting of solid, its temperature does not change.

During the melting of solid, all the heat given to the solid spend in the change of state from solid to liquid. That's why the temperature of the material does not change.
Question 7
1 / -0

Among the following reactions, the impossible one is :

A
\({ }^{2} \mathrm{He}_{4}+{ }^{4} \mathrm{Be}_{9} \rightarrow{ }^{0} \mathrm{n}_{1}+{ }^{6} \mathrm{C}_{12}\)

B
\({ }^{2} \mathrm{He}_{4}+{ }^{7} \mathrm{~N}_{14} \rightarrow{ }^{1} \mathrm{H}_{1}+{ }^{8} \mathrm{O}_{17}\)

C
\(4\left({ }^{1} \mathrm{H}_{1}\right) \rightarrow{ }^{2} \mathrm{He}_{4}+{ }^{2}\left({ }^{-1} \mathrm{e}_{0}\right)\)

D
\({ }^{3} \mathrm{Li}_{7}+{ }^{1} \mathrm{H}_{1} \rightarrow{ }^{4} \mathrm{Be}_{8}\)

Solution

For possible reactions :

Sum of mass numbers LHS should be equal to the sum of mass numbers on RHS of the equation.

As well as sum of Atomic numbers on LHS should be equal to the sum of Atomic numbers on RHS of the equation.

So, in option (C) we can see that on LHS mass number is \(4 \times 1=\) 4 and Atomic number is \(4 \times 1=4\). on RHS mass number is \(2+\) \(2(-1)=0\) and Atomic number is \(4+0=4\).

So, mass number are different in LHS and RHS of the equation.So, this reaction is not possible.
Question 8
1 / -0

When the tension in a metal wire is \(T_{1}\), its length is \({l}_{{t}}\). When the tension is \({T}_{2}\), its length is \(l_{2}\). The natural length of wire is:

A
\(\frac{{T}_{2}}{{~T}_{1}}\left({l}_{1}+{l}_{2}\right)\)

B
\({T}_{1} {l}_{1}+{T}_{2} {l}_{2}\)

C
\(\frac{l_{1} T_{2}-l_{2} T_{1}}{T_{2}-T_{1}}\)

D
\(\frac{l_{1} T_{2}-l_{2} T_{1}}{T_{2}+T_{1}}\)

Solution

When tension \({T}_{1}\) then length \(=l_{1}\)

When tension \({T}_{2}\) then length \(={l}_{2}\)

\(\Delta {l}=\frac{{Fl}}{{AY}}=\frac{\Delta {l}_{1}}{\Delta {l}_{2}}=\frac{{F}_{1}}{{~F}_{2}}=\frac{{T}_{1}}{{~T}_{2}}\)

\(\frac{l_{1}-l}{l_{2}-l}=\frac{{T}_{1}}{{T}_{2}}\)

\(\Rightarrow l_{1} {T}_{2}-l_{1} {T}_{2}=l_{2} {T}_{1}-lT_{1}\)

Natural length of wire is \(=\frac{l_{2} T_{1}-l_{1} T_{2}}{T_{2}-T_{1}}\)
Question 9
1 / -0

At t = 0, an arrow is fired vertically upwards with a speed of 98 ms-1. A second arrow is fired vertically upwards with the same speed at t = 5 s. Then,

A
the two arrows will be at the same height above the ground at t = 10 s

B
the two arrows will reach back to their starting points at t = 20 s and t = 30 s

C
the ratio of the speed of the first arrow and that of the second arrow at t = 20 s will be 2 : 1

D
the maximum height attained by either arrow will be 980 m

Solution

\(\begin{aligned} & \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ & \Rightarrow 98 \mathrm{t}-\frac{1}{2} 9.8 \mathrm{t}^2=98(\mathrm{t}-5)-\frac{1}{2} 9.8(\mathrm{t}-5)^2 \\ & \Rightarrow \mathrm{t}^2-(\mathrm{t}-5)^2-100=0 \\ & \Rightarrow 10 \mathrm{t}=125 \\ & \mathrm{t}=12.5 \mathrm{~s}\end{aligned}\)

Now, total time for first arrow will be

\(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\)

\(\Rightarrow 98 \mathrm{t}-\frac{1}{2} 9.8 \mathrm{t}^2=0\)

\(t(t-20)=0\)

\(\mathrm{t}_1=20 \mathrm{~s}\)

\(\Rightarrow \mathrm{t}_2=25 \mathrm{~s}\)

(Now at \(\mathrm{t}=20 \mathrm{~s}\) )

\(\mathrm{v}=\mathrm{u}+\text { at }\)

\(\mathrm{v}_1=98 \quad \mathrm{v}_2=98 \frac{1}{2}\)

\(\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{2}{1}\)

\(\mathrm{v}_1: \mathrm{v}_2=2: 1\)
Question 10
1 / -0

The work function of Cesium is 2.27eV. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600nm wavelength is:

A
0.00206785 eV

B
-0.206785 eV

C
0.206785 eV

D
2.06785 eV

Solution

Given:

\(\lambda=600 \mathrm{~nm}\)

We know that

\(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

\(h=4.1357 \times 10^{-15} \mathrm{eV}\)

The energy of the photon with a wavelength lambda is given by \(E=\frac{h c}{\lambda}\)

Where \(h\) is the planck's constant

\(c\) is the speed of light

\(\lambda\) is the wavelength

Put the values in formula given above.

\(E=\frac{4.1357 \times 10^{-15} \times 3 \times 10^{8} }{600 \times 10^{-9}}\)

\(=\frac{4.1357 \times 3}{6}\)

\(=2.06785 \mathrm{~eV}\)

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Physics Test - 28

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Physics Test - 28

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Question 1

\(0.1\) poise

\(0.5\) poise

\(0.7\) poise

\(0.9\) poise

Solution

Question 2

\(\frac{2 \lambda_{1}}{\lambda_{2}}\)

\(\frac{2 \lambda_{2}}{\lambda_{1}}\)

\(\frac{\lambda_{1}}{2 \lambda_{2}}\)

\(\frac{\lambda_{2}}{2 \lambda_{1}}\)

Solution

Question 3

Deuteron

Proton

Neutron

\(\beta\) - particle

Solution

Question 4

\(12.89 \mathrm{~g} / \mathrm{cm}^{3}\)

\(14 \mathrm{~g} / \mathrm{cm}^{3}\)

\(11.69 \mathrm{~g} / \mathrm{cm}^{3}\)

Zero

Solution

Question 5

\(\frac{1}{\left(\gamma_{2}-\gamma_{1}\right) \Delta t}\)

\(\frac{1}{\left(\gamma_{1}-\gamma_{2}\right) \Delta t}\)

\(\left(\gamma_{1}-\gamma_{2}\right) \Delta t\)

\(\left(\gamma_{2}-\gamma_{1}\right) \Delta t\)

Solution

Question 6

May increase or decrease depending on the nature of solid

Decreases

Increases

Does not change

Solution

Question 7

\({ }^{2} \mathrm{He}_{4}+{ }^{4} \mathrm{Be}_{9} \rightarrow{ }^{0} \mathrm{n}_{1}+{ }^{6} \mathrm{C}_{12}\)

\({ }^{2} \mathrm{He}_{4}+{ }^{7} \mathrm{~N}_{14} \rightarrow{ }^{1} \mathrm{H}_{1}+{ }^{8} \mathrm{O}_{17}\)

\(4\left({ }^{1} \mathrm{H}_{1}\right) \rightarrow{ }^{2} \mathrm{He}_{4}+{ }^{2}\left({ }^{-1} \mathrm{e}_{0}\right)\)

\({ }^{3} \mathrm{Li}_{7}+{ }^{1} \mathrm{H}_{1} \rightarrow{ }^{4} \mathrm{Be}_{8}\)

Solution

Question 8

\(\frac{{T}_{2}}{{~T}_{1}}\left({l}_{1}+{l}_{2}\right)\)

\({T}_{1} {l}_{1}+{T}_{2} {l}_{2}\)

\(\frac{l_{1} T_{2}-l_{2} T_{1}}{T_{2}-T_{1}}\)

\(\frac{l_{1} T_{2}-l_{2} T_{1}}{T_{2}+T_{1}}\)

Solution

Question 9

the two arrows will be at the same height above the ground at t = 10 s

the two arrows will reach back to their starting points at t = 20 s and t = 30 s

the ratio of the speed of the first arrow and that of the second arrow at t = 20 s will be 2 : 1

the maximum height attained by either arrow will be 980 m

Solution

Question 10

0.00206785 eV

-0.206785 eV

0.206785 eV

2.06785 eV

Solution

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