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Physics Class 12th System of Particles and Rotational Motion physics ncert exemplar solutions class 12th chapter two

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.

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Vishal Baghel | Contributor-Level 10

2 months ago

A/c to question

f + 76 = 2f

f = 76

last frequency

= f + 76 = 2 × 76

= 152 H₂

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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)

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

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

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:

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 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

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:

alok kumar singh

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

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

a = 5

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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.

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