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

6 months ago

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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Similar Questions for you

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)

Alok Kumar Singh

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= $50 \times \frac{3}{5} = 30 N$

Therefore, NA = $F_{ω} = 30 N$ - (i)

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

Alok Kumar Singh

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

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

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

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

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

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

A set of 20 tuning forks is arranged in series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is ___________Hz.

Vishal Baghel

A/c to question

f + 76 = 2f

f = 76

last frequency

= f + 76 = 2 × 76

= 152 H₂

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is-

Vishal Baghel

K.ER (Rotational) = $\frac{1}{2} I_{C M} ω^{2} = \frac{1}{2} \times \frac{2}{5} M R^{2} ω^{2} = \frac{M R^{2} ω^{2}}{5}$ - (1)

$K . E_{T o t a l} = K . E_{R} + K . E_{T}$

$= \frac{M R^{2} ω^{2}}{5} + \frac{1}{2} M V_{c m^{2}}$

$= \frac{7 M R^{2} ω^{2}}{10}$ - (2)

$\frac{k . E_{R}}{K . E_{T o t a l}} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$

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

Payal Gupta

At maximum height velocity (v) = 0

So, momentum = mv = m × 0 = 0

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