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System of Particles and Rotational Motion

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2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be

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Payal Gupta

Contributor-Level 10

mg – T = ma. (1)

T * R = l α. (2)

a = α R. (3)

With the help of equations (1), (2) and (3), we get

$a = \frac{m g}{m + \frac{l}{R^{2}}}$

$v = \sqrt[]{2 a h} = ω R$

$\Rightarrow ω^{2} = \frac{2 m g h}{l + m R^{2}}$

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New answer posted

2 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ? = ?0 $(1 - \frac{x^{2}}{L^{2}}) k g / m$ and length L (in meter) is $\frac{3 L}{α} m .$ The value of is _________. (where x is the distance from end A)

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Vishal Baghel

Contributor-Level 10

Linear mass density, $λ = λ_{0} (1 - \frac{x^{2}}{L^{2}}) k g / m$

$M = \int λ d x = \int_{0}^{L} λ_{0} (1 - \frac{x^{2}}{L^{2}}) d x$

$= λ_{0} (L - \frac{L^{3}}{3 L^{2}}) = \frac{2 L λ_{0}}{3}$

$A s, x_{C} = \frac{3 L}{α} = \frac{9}{8} L . \Rightarrow α = \frac{3 * 8}{9} = 8 / 3$

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

The position vector of 1kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) m s^{- 1} .$ The magnitude of is angular momentum is $\sqrt[]{x} N m$ where x is________.

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Payal Gupta

Contributor-Level 10

$\vec{L} = \vec{r} * m \vec{v}$

= $(3 \hat{i} - \hat{j}) * (3 \hat{j} + \hat{k})$

$= 9 \hat{k} + 3 (- \hat{j}) - \hat{i}$

$= - \hat{i} - 3 \hat{j} + 9 \hat{k}$

$| \vec{L} |$ $=$ =√ $1 + 9 + 81$ $=$ =√91

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Match List – I with List – II

	List – I		List – II
A.	Moment of inertia of solid sphere of radius R about and tangent.	I.	5/3MR²
B.	Moment of inertia of hollow sphere of radius (R) about any tanget	II.	7/5MR²
C.	Moment of inertia of circular ring of radius (R) about its diameter.	III.	1/4MR²
D.	Moment of inertial of circular disc of radius (R) about any diameter.	IV.	1/2MR²

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Use formula for M.I.

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is . The difference of resultant intensities, at the point A and B, will be_________I.

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Vishal Baghel

Contributor-Level 10

$I_{A} = 9 l + 4 l + 2 \sqrt[]{9 l * 4 l} c o s 0 °$

$= 13 l + 12 l = 25 l$

$I_{B} = 9 l + 4 l + 2 \sqrt[]{9 l * 4 l} c o s π$

$= 13 l - 12 l = l$

$∴ I_{A} - I_{B} = 24 l$

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An object is projected in the air with initial velocity u at an angle q. The projectile motion is such that the horizontal range R, in maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be_______degree.

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Vishal Baghel

Contributor-Level 10

$R = \frac{u^{2} s i n (2 * 45 °)}{g} = \frac{u^{2}}{g}$

$\frac{R}{2} = \frac{u^{2}}{2 g} = \frac{u^{2} s i n 20}{g}$

$s i n 2 θ = \frac{1}{2}$

$\Rightarrow 2 θ = 30 ° \Rightarrow θ = 15 °$

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Two bodies of mass 1kg and 3kg have position vectors $\hat{i} + 2 \hat{j} + \hat{k} a n d - 3 \hat{i} - 2 \hat{j} + \hat{k}$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector:

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Vishal Baghel

Contributor-Level 10

$m_{1} = 1 k g, {\vec{r}}_{1} = \hat{i} + 2 \hat{j} + \hat{k}$

$m_{1} = 3 k g, {\vec{r}}_{2} = - 3 \hat{i} - 2 \hat{j} + \hat{k}$

${\vec{r}}_{c o m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}} = \frac{1}{4} (\hat{i} + 2 \hat{j} + \hat{k} + - 9 \hat{i} - 6 \hat{j} + 3 \hat{k})$

= $2 \hat{i} - \hat{j} + \hat{k}$

$| {\vec{r}}_{c o m} | = \sqrt[]{4 + 1 + 1} = \sqrt[]{6 .}$

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A rolling wheel of 12kg is on an inclined plane at position P and connected to a mass of 3kg through a string of fixed length and pulley as shown in figure.

Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $\frac{1}{2} \sqrt[]{x g h} m / s .$ The value of x is…………

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Payal Gupta

Contributor-Level 10

By conservation of Energy

$m g h = \frac{1}{2} l_{P} ω^{2}$

3gh $= \frac{1}{2} (M R^{2} + M R^{2}) ω^{2}$

$3 g h = m v^{2}$

$v^{2} = \frac{3 g h}{12}$

$v = \frac{\sqrt[]{g h}}{2} = \frac{\sqrt[]{x g h}}{2}$

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ , respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ then the value of $α$ is:

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alok kumar singh

Contributor-Level 10

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

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New answer posted

3 months ago

System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt[]{x (h^{2} - g L)} .$ The value of x is

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Payal Gupta

Contributor-Level 10

Conservation of Energy b/w (1) & (2)

$\frac{1}{2} m v^{2} + m g l = \frac{1}{2} m u^{2}$

$v = \sqrt[]{u^{2} - 2 g l}$

$\vec{Δ v} = v_{\hat{j}} - u_{\hat{i}}$

$| Δ \vec{v} | = \sqrt[]{{(u)}^{2} + {(\sqrt[]{u^{2} - 2 g l})}^{2}}$

$= \sqrt[]{u^{2} + u^{2} - 2 g l}$

$= \sqrt[]{2} (u^{2} - g l)$

x = 2

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