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8 months ago

Physics Class 11th

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.

Reason R : On heating, the length of the rod increases.

In the light of the above statements, choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Thermal stress is developed on heating when expansion of rod is hindered.

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8 months ago

Physics Units and Measurement

The pitch of the screw gauge is 1mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72^nd division on circular scale coincides with the reference line, The radius of the wire is:

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Vishal Baghel

Contributor-Level 10

$L C = \frac{1}{100} m m = 0.01 m m$

Zero error = +0.08 mm.

Diameter = 1 + 72 * 0.01 – 0.08 = 1.64 mm

Radius = 0.82 mm

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8 months ago

Physics Class 12th

Zener breakdown occurs in a P- n junction having P and n both:

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Vishal Baghel

Contributor-Level 10

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

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8 months ago

Physics Units and Measurement

The period of oscillation of a simple pendulum is $T = 2 π \sqrt[]{\frac{L}{g}} .$ Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be:

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{L}{g}} o r T^{2} = 4 π^{2} (\frac{L}{g})$

$\Rightarrow g = 4 π^{2} (\frac{L}{T^{2}})$

$\Rightarrow \frac{Δ g}{g} % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) % = [\frac{1 m m}{1 m} + \frac{2 * 0.01}{1.95}] * 100 % = (0.001 + 0.0102) * 100 = 1.13 %$

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8 months ago

Physics Class 12th

Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current 'i' just after switch 'S' is closed will be:

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Vishal Baghel

Contributor-Level 10

R = $2 Ω$

L = 2 mH

E = 9V

$i = \frac{ε}{2 R} = \frac{9 v}{4 Ω} = 2.25 A$

Just after the switch 'S' is closed, the inductor acts as open circuit.

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8 months ago

Physics Class 11th

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

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Vishal Baghel

Contributor-Level 10

According to KTG, the gas exerts pressure because its molecule :

Suffer change in momentum when impinge on the walls of container.

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8 months ago

Physics Class 11th

Four identical particles of equal masses 1kg made to move along the circumference of a circle of radius 1m under the actin of their own mutual gravitational attraction. The speed of each particle will be:

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Vishal Baghel

Contributor-Level 10

$F_{1} c o s 45 ° + F_{2} c o s 45 ° + F_{3} = \frac{m v^{2}}{r}$

$\Rightarrow \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(2 r)}^{2}} = \frac{m v^{2}}{r}$

$\Rightarrow v = \sqrt[]{\frac{G m}{4 r} (2 \sqrt[]{2} + 1)}$

Putting m = 1 kg and r = 1 m,

$v = \frac{1}{2} \sqrt[]{G (1 + 2 \sqrt[]{2})}$

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8 months ago

Physics Oscillations

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. New amplitude of oscillation will be:

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Vishal Baghel

Contributor-Level 10

Velocity of block in equilibrium, in first case,

$v = A ω = A . \sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v' = A' ω' = A' \sqrt[]{\frac{k}{M + m}}$

From conservation of momentum,

Mv = (M + m) v'

$\Rightarrow M A \sqrt[]{\frac{k}{M}} = (M + m) A' \sqrt[]{\frac{k}{M + m}} \Rightarrow A' = A \sqrt[]{\frac{M}{M + m}}$

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8 months ago

Physics Units and Measurement

The work done by a gas molecule in an isolated system is given by $W = α β^{2} e^{- \frac{x^{2}}{α k T}}$ , where x is the displacement, k is the Boltzmann constant and T is the temperature. α and β are constants. Then the dimensions of β will be:

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Vishal Baghel

Contributor-Level 10

Since, $\frac{x^{2}}{α k T}$ $\frac{^{}}{}$ should be dimensionless.

So, dimension of $α, [α] = \frac{L^{2}}{M L^{2} T^{- 2}} = M^{- 1} T^{2}$ $\frac{^{}}{^{}^{}}^{}^{}$

Dimension of $α β^{2}$ $^{}$ should be that of W.

So, $[α β^{2}] = M L^{2} T^{- 2}$ $^{}^{}^{}$

$\Rightarrow [β^{2}] = \frac{M L^{2} T^{- 2}}{M^{- 1} T^{2}} = M^{2} L^{2} T^{- 4} \Rightarrow [β] = M L T^{- 2}$

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8 months ago

Physics Class 12th

A current through a wire-depends on time as

i = α₀t + βt²

Where α₀ = 20A/s and β = 8As^-². Find the charge crossed through a section of the wire in 15s.

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Vishal Baghel

Contributor-Level 10

$\frac{d q}{d t} = t = 20 t + 8 t^{2}$

$\Rightarrow \int_{0}^{q} d q = \int_{0}^{15} (20 t + 8 t^{2}) d t$

$\Rightarrow q = 20 * \frac{1 5^{2}}{2} + 8 . \frac{1 5^{3}}{3} = 11250 C$

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