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

Option 1 - 3

Option 2 - 2

Option 3 - 1

Option 4 - 5

2 Views|Posted 8 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

8 months ago

Correct Option - 2

Detailed Solution:

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

Upvote

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

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering