Physics NCERT Exemplar Solutions Class 12th Chapter Nine Laws of Motion Class 12th Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination , is given by:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mi>L</mi> <mo>/</mo> <mrow> <mo>(</mo> <mrow> <mi>g</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mi>L</mi> <mo>/</mo> <mrow> <mo>(</mo> <mrow> <mi>g</mi> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mi>L</mi> <mo>/</mo> <mi>g</mi> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mi>L</mi> <mo>/</mo> <mi>g</mi> <mrow> <mo>(</mo> <mrow> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>α</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

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Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 1

Detailed Solution:

goff = g cos

$∴ T = 2 π \sqrt[]{\frac{L}{g c o s α}}$

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A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1cm on main scale and 50 Vernier scale divisions are equal to 49 main scale division, then least count of the travelling microscope is __________× 10^-6 m.

Let ‘x’ be the value of one division on main scale

$x = \frac{1}{40} c m$

Let ‘y’ be the value of one division on vernier

Now given

50y = 49 x

$y = \frac{49 x}{50}$

$\begin{array}{l} = 5 \times 1 0^{- 4} c m \\ = 5 \times 1 0^{- 6} m \\ 5 \end{array}$

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

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In the arrangement shown in figure a₁, a₂, a₃ and a₄ are the accelerations of masses m₁, m₂, m₃ and m₄ respectively. Which of the following relation is true for this arrangement?

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A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

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A block of metal weighing 2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet relasing water at a rate of 1 kgs^-1 and at a speed of 10 ms^-1. Then, the initial acceleration of the block, in ms^-2, will be:

Alok Kumar Singh

$a = \frac{F}{m} = \frac{m v}{m} = \frac{1 * 10}{2} = 5 m / s e c^{2}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine 2025

Physics NCERT Exemplar Solutions Class 12th Chapter Nine 2025

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