AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A uniform disc can rotate about a fixed horizontal axis passing through its center and perpendicular to its plane. A light chord is wrapped around the rim of the disc (mass 2 kg and radius 1 m) and mass of 1 kg is tied to the free end. If it is released from rest,
Column – I Column – II
(a) The tension in the chord is ----N. (p) 10
(b) In the first 4 second the angular displacement of the disc is---rad (q) 5
(c) The work done by the torque on the disc in the first 4 second is ----J. (r) 40
(d) Acceleration of mass is----- $m / s^{2}$ (s) 200
(t) 50

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(a) \to (q),_(b) \to (s), (c) \to (r), (d) \to (p)$

$(a) \to (q), (b) \to (r), (c) \to (s), (d) \to (q)$

$(a) \to (r), (b) \to (q), (c) \to (s), (d) \to (p)$

$(a) \to (p), (b) \to (r), (c) \to (s), (d) \to (q)$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

By FDB of particles

$m g - T = m a$

By FBD of disc

$T R = I α = I \frac{a}{R} \Rightarrow T = \frac{M R^{2}}{2} . \frac{a}{R^{2}} \Rightarrow \frac{M a}{2} = \frac{2 a}{2} = a$

$\begin{array}{l} m g - a = m a \\ (1) (10) = a + 1 (a) \\ a = 5 m / s^{2} \end{array}$

$∴ a = 5 m s^{- 2}$ T=5N

$α = \frac{a}{R} = 5 rad s^{- 2}$

$θ = ω t + \frac{1}{2} α t = \frac{1}{2} \times 5 \times 16 = 40 r a d$

$W = τ Δ θ = 5 \times 40 = 200 J$

$Δ K E = W = 200 J$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!