Physics Test

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Physics Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of S.I. to C.G.S. units for Stefan's constant is :

A
1100

B
11000

C
100

D
1000

Solution

Units of stefan's constant \(\sigma=W m^{-2} K^{-4}=k g m^{0} s^{-3} K^{-4}\) In CGS Units, \(1 kg =1000 gm\)
\(\therefore\) units of stefan's constant in CGS Units \(=1000 gs ^{-3} K ^{-4}\)
Therefore, Ratio of S.I to C.G.S units \(=1000\)
Hence, option D is correct.
Question 2
1 / -0

In a particular system the units of length, mass and time are chosen to be 10 cm, 10 g, and 0.1 s respectively. The units of force in this system will be equal to ?

A
0.1N

B
1N

C
10N

D
100N

Solution

1 newton \(=1 kgm / s ^{2}\)
1 \(kg =10^{3}\) and \(1 m =10^{2} cm\)
1 \(N=\frac{\left(10^{3} g\right)\left(10^{2} cm \right)}{s^{2}}\)
1 \(N=\frac{100 \times(10 g ) \times 10(10 cm )}{100 \times(0.1 s)^{2}}\)
\(=10 \times \frac{(10 g )(10 cm )}{(0.1 s)^{2}}\)
\(1 N=10 \times\) New unit of force
Thus, New unit \(=1 / 10=0.1 N\)
Hence option A is the correct answer.
Question 3
1 / -0

Young's modulus of steel is 2.0×10¹¹N/m². Express it in dyne/cm².

A
2.0×10¹²dyne/cm²

B
2.0×10¹¹dyne/cm²

C
4.0×10¹²dyne/cm2

D
4.0×10¹¹dyne/cm²

Solution

Dyne is the CGS unit of Force.
Dimension of force is \(\left[\mathrm{MLT}^{-2}\right]\) Thus, dimension of Young's modulus is \(|\mathrm{F} / \mathrm{A}|=\mathrm{ML}^{-1} \mathrm{T}^{-2}\)
\(1 \mathrm{kg}=1000 \mathrm{g}, 1 \mathrm{m}=100 \mathrm{cm}\)
Thus, \(2 \times 10^{11} \mathrm{N} / \mathrm{m}^{2}=2 \times 10^{11} \times(1000 \mathrm{g}) \times(100 \mathrm{cm})^{-1} \times 1 \mathrm{s}^{-2}\)
\(\quad=2 \times 10^{11} \times 10\)
\(=2 \times 10^{12} \mathrm{dyne} / \mathrm{cm}^{2}\)
Hence option A is the correct answer.
Question 4
1 / -0

A scientist proposes a new temperature scale in which the ice point is 25X (X is the new unit of temperature) and the steam point is 305X. The specific heat capacity of water in this new scale is (in Jkg⁻¹X^-1):

A
4.2×10³

B
3.0×10³

C
1.2×10³

D
1.5×10³

Solution

Specific heat capacity of water, \(S=4200 \frac{J}{k g \times K}\)
Given : For new temperature scale, Ice point \(=25 X\) Steam point \(=305 X\)
Difference in the steam point and ice point is equal to \(100 K\) \(\therefore 305 X-25 X=100 K\)
\(\Longrightarrow 1 K=2.8 X.\)
This means 1 Kelvin is equal to \(2.8 X\)
\(\therefore S=4200 \frac{J}{k g \times K}=4200 \frac{J}{k g \times 2.8 X}=1500 \frac{J}{k g \times X}\)
\(\Longrightarrow\) Specific heat capacity of water is \(1.5 \times 10^{3} J kg ^{-1} X ^{-1}\)
Hence option D is the correct answer.
Question 5
1 / -0

The density of a material in CGS system of units is 4g/cm³. In a system of units in which unit of length is 10cm and unit of mass is 100g, the value of density of material will be :

A
0.4 Unit

B
40 Unit

C
4400 Unit

D
0.04 Unit

Solution

The density of material is given as \(d=4 g / c m^{3}\)
In another system, \(1 cm =0.1\) unit and \(1 g =0.01\) unit
Thus density of material in that system \(d=4 \times \frac{0.01}{(0.1)^{3}}=40\) Unit.
Hence option B is the correct answer.
Question 6
1 / -0

Dimensions of specific heat are :

A
[ML²T⁻²K]

B
[ML²T⁻²K⁻¹]

C
[ML²T²K⁻¹]

D
[L²T⁻²K⁻¹]

Solution

The specific heat \(s\) is the amount of heat \(Q\) per unit mass \(m\) required to raise the temperature \(\theta\) by one degree Celsius.
So, \(s=\frac{Q}{m \theta}=\frac{M L^{2} T^{-2}}{M K}=\left[L^{2} T^{-2} K^{-1}\right]\)
Hence option D is the correct answer.
Question 7
1 / -0

The physical quantity that has no dimensions is:

A
angular velocity

B
linear momentum

C
angular momentum

D
strain

Solution

Angular velocity \(=M^{0} L^{0} T^{-1}\) Linear momentum \(=m v=M^{1} L^{1} T^{-1}\)
Angular momentum \(=m v r=M^{1} L^{2} T^{-1}\)
Strain \(=\frac{\Delta L}{L}=\frac{M^{0} L^{1} T^{0}}{M^{0} L^{1} T^{0}}=M^{0} L^{0} T^{0}\)
Hence option D is the correct answer.
Question 8
1 / -0

The dimensions of resistivity in terms of M, L, T and Q, where Q stands for the dimensions of charge is :

A
ML³T⁻¹Q⁻²

B
ML³T⁻²Q⁻¹

C
ML²T⁻¹Q⁻¹

D
MLT⁻¹Q⁻¹

Solution

\(R=\frac{\rho l}{A}\)
Power \(=I^{2} R\) Dimension of \(R\) is \(\frac{M L^{2} T^{-1}}{Q^{2} T^{-2}}=M L^{2} T Q^{-2}\)
Now by equation, \(\rho=\frac{R A}{L}=M L^{2} T Q^{-2} L=M L^{3} T Q^{-2}\)
Hence, option \(A\) is correct.
Question 9
1 / -0

Modulus of Elasticity is dimensionally equivalent to:

A
Stress

B
Surface tension

C
Strain

D
Coefficient of viscosity

Solution

We know that,
The modulus of elasticity is the ratio of the stress and strain.
\(Y=\frac{\text {stress}}{\text {strain}}\)
Strain is unit less.
Therefore,
The dimension formula of modulus of elasticity is
\(Y=\frac{M L^{-1} T^{-2}}{u \text { nitless }}\)
\(Y=M L^{-1} T^{-2}\)
We know that,
The stress is the force upon area.
Stress \(=\frac{F}{A}\)
The dimension formula of stress
Stress \(=M L^{-1} T^{-2} \ldots .(1)\)
From eqyation (I) and (II)
\(Y=M L^{-1} T^{-2}=\) stress
Hence, The dimension formula of modulus of elasticity is equivalent to
stress.Hence, option \(A\) is correct.
Question 10
1 / -0

Dimensional formula for capacitance is:

A
M⁻¹L⁻²T⁴I²

B
M¹L²T⁴I⁻²

C
M¹L²T²

D
MLT⁻¹

Solution

\(E=\frac{Q^{2}}{2 C}\)
Dimensions of \(C\) will be dimension of \(\frac{Q^{2}}{E}=\frac{I^{2} T^{2}}{M L^{2} T^{-2}}=M^{-1} L^{-2} T^{4} I^{2}\)
Hence, option \(A\) is correct.

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Physics Test - 6

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Physics Test - 6

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Question 1

1100

11000

100

1000

Solution

Question 2

0.1N

1N

10N

100N

Solution

Question 3

2.0×1012dyne/cm2

2.0×1011dyne/cm2

4.0×1012dyne/cm2

4.0×1011dyne/cm2

Solution

Question 4

4.2×103

3.0×103

1.2×103

1.5×103

Solution

Question 5

0.4 Unit

40 Unit

4400 Unit

0.04 Unit

Solution

Question 6

[ML2T−2K]

[ML2T−2K−1]

[ML2T2K−1]

[L2T−2K−1]

Solution

Question 7

angular velocity

linear momentum

angular momentum

strain

Solution

Question 8

ML3T−1Q−2

ML3T−2Q−1

ML2T−1Q−1

MLT−1Q−1

Solution

Question 9

Stress

Surface tension

Strain

Coefficient of viscosity

Solution

Question 10

M−1L−2T4I2

M1L2T4I−2

M1L2T2

MLT−1

Solution

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2.0×10¹²dyne/cm²

2.0×10¹¹dyne/cm²

4.0×10¹²dyne/cm2

4.0×10¹¹dyne/cm²

4.2×10³

3.0×10³

1.2×10³

1.5×10³

[ML²T⁻²K]

[ML²T⁻²K⁻¹]

[ML²T²K⁻¹]

[L²T⁻²K⁻¹]

ML³T⁻¹Q⁻²

ML³T⁻²Q⁻¹

ML²T⁻¹Q⁻¹

MLT⁻¹Q⁻¹

M⁻¹L⁻²T⁴I²

M¹L²T⁴I⁻²

M¹L²T²

MLT⁻¹