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Physics NCERT Exemplar Solutions Class 12th Chapter Seven

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3 months ago

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

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

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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Physics NCERT Exemplar Solutions Class 12th Chapter Seven Alternating Current

An alternating emf E = 440 sin 100t is applied to a circuit containing an inductance of $\frac{\sqrt[]{2}}{π} H .$ If an a.c. ammeter is connected in the circuit, its reading will be:

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Vishal Baghel

Contributor-Level 10

E = 440 sin 100 t $ω = 100 π$

$L = \frac{\sqrt[]{2}}{π} H .$ $X_{L} = ω L = 100 π . \frac{\sqrt[]{2}}{π} = 100 \sqrt[]{2} Ω .$

$= \frac{220}{100} = 2.2 A$

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

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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Physics NCERT Exemplar Solutions Class 12th Chapter Seven Physics Alternating Current

A series $L - R$ circuit is connected to a battery of emf $V$ . If the circuit is switched on at $t =$ 0 , then the time at which the energy stored in the inductor reaches $(1 / n)$ times of its maximum value, is :

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alok kumar singh

Contributor-Level 10

Sol. Mono atomic $? C_{v} = \frac{3 R}{2} C_{p} = \frac{5 R}{2}$

Di- atomic $? C_{V} = \frac{5 R}{2} C_{p} = \frac{7 R}{2}$

(Rigid)

Di-atomic $? C_{v} = \frac{7 R}{2} C_{p} = \frac{9 R}{2}$

(Non-Rigid)

Tri-atomic $? C_{V} = 3 R C_{P} = 4 R$

(Rigid)

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Physics NCERT Exemplar Solutions Class 12th Chapter Seven Semiconductors

Identify the operation performed by the circuit given below:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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

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

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O^{'}$ (corner point) is:

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alok kumar singh

Contributor-Level 10

Sol. $m (L) = m_{1} S_{1} (Δ T)$

$\Rightarrow m (3.4 * 10^{5}) = (200) (4200) (25)$

$\Rightarrow m = 61.7$

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

A circular coil has moment of inertia $0.8 k g m^{2}$ around any diameter and is carrying current to produce a magnetic moment of $20 {A m}^{2}$ . The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 T$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{?}$ will be:

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alok kumar singh

Contributor-Level 10

Sol. $K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

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

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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