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Physics NCERT Exemplar Solutions Class 12th Chapter Seven

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2 months ago

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

A $\sqrt[]{34} m$ long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w / F_f will be:

(Use g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

mg = 10 * 10

= 100 of

NA = $F_{ω}$ reaction force offered by the wall

$\sqrt[]{N_{3}^{2} + f_{B}^{2}} = F_{f}$ reaction force offered by the floor.

By NLM 1

Translational fB = NA

NB = mg = 100N

Ac² + Bc² = AB²

Ac² = 34 – 9

Ac = 5m

Rotational equilibrium $τ_{B} = 0$

mg $\frac{l}{2} c o s θ = N_{A} l s i n θ$

$100 \frac{c o s θ}{2} = N_{A} s i n θ$

NA = 50 cot

Now cot = $\frac{3}{5}$

N_A= $50 * \frac{3}{5} = 30 N$

Therefore, NA = $F_{ω} = 30 N$ - (i)

$F_{f} = \sqrt[]{N_{B}^{2} + f_{B}^{2}} = \sqrt[]{10 0^{2} + 3 0^{2}} = 10 \sqrt[]{109} N$ - (2)

$\frac{f_{ω}}{F_{f}} = \frac{30}{10 \sqrt[]{109}} = \frac{3}{109}$

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2 months ago

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

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

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alok kumar singh

Contributor-Level 10

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

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2 months ago

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

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

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Payal Gupta

Contributor-Level 10

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

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2 months ago

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

For a series LCR circuit, I vs curve is shown:

Choose the most appropriate answer from the options given below:

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alok kumar singh

Contributor-Level 10

At w_r impedance of the circuit is equal to the resistance of the circuit.

Left of w_r, circuit is mainly capacitive

Right of w_r, the circuit is mainly Inductive.

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

A spherical of 1 kg mass and radius R is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^{2} ω .$ The value of a will be:

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alok kumar singh

Contributor-Level 10

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

$\frac{5}{3} * 1 R^{2} ω = \frac{5}{3} R^{2} ω = \frac{a}{3} R^{2} ω$ =

a = 5

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2 months ago

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

A spherical of 1 kg mass and radius R is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^{2} ω .$ The value of a will be

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Payal Gupta

Contributor-Level 10

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

= $\frac{5}{3} * 1 R^{2} ω = \frac{5}{3} R^{2} ω = \frac{a}{3} R^{2} ω$

a = 5

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Seven Alternating Current

The effective current I in the given circuit at very high frequencies will be _________A.

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Vishal Baghel

Contributor-Level 10

At very high frequencies

$X_{C} = \frac{1}{ω C} \to 0$

$X_{L} = ω L \to \infty$

Thus equivalent circuit

$Z = 1 + 2 + 2 = 5 Ω$

$l = \frac{220}{5} = 44 A$

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2 months ago

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

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the centre will be________m.

Given, the length of the rod is $10 \sqrt[]{3} m .$

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alok kumar singh

Contributor-Level 10

$k = \sqrt[]{\frac{l}{m}} = \sqrt[]{\frac{m l^{2} / 12}{m}} = \frac{l}{2 \sqrt[]{3}}$

$k = \frac{10 \sqrt[]{3}}{2 \sqrt[]{3}} = 5$

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2 months ago

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

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Vishal Baghel

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 * 3 R}}$

$x = \frac{g}{2} = 5$

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2 months ago

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

As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant 2Nm^‑1. If both are given velocity $υ$ in opposite directions, then maximum elongation of the spring is:

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Vishal Baghel

Contributor-Level 10

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_{r} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{m m}{m + m} = \frac{m}{2}$

By wet

$w_{S P} = Δ K . E .$

$- \frac{1}{2} k x^{2} = 0 - \frac{1}{2} (\frac{m}{2}) {(2 v)}^{2}$

$\frac{2}{2} x^{2} = 0.25 v^{2}$

$x^{2} = 0.25 v^{2}$

$x = \frac{v}{2}$

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