A uniform disc of mass 5 kg and radius 1 m is placed vertically on a rough horizontal surface. A horizontal time varying force is applied on the center of disc and its variation with time is shown in the graph. The co–efficient of friction between disc and surface is 0.4. (Take g = 10 m / $s^{2}$ )
Then which of the following option is/are correct

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

at t = 7 sec disc is rolling without slipping

at t = 4 sec disc is rolling without slipping

at t = 3 sec disc is rolling without slipping

at t = 3 second friction force is 10 N

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$\begin{array}{l} T r a n s l a t i o n : F - f = m a - - - (1) \\ \Rightarrow R o t a t i o n : τ = I α \\ \Rightarrow f R = \frac{m R^{2}}{2} \frac{a}{R} (∵ a = R α) \\ \Rightarrow f = \frac{m a}{2} - - - (3) \end{array}$

$\begin{array}{l} f r o m (1) & (3) \\ \Rightarrow f = \frac{F}{3} = \frac{10 t}{3} \\ (∵ \frac{F}{t} = \frac{80}{8} \Rightarrow F = 10 t) \end{array}$

$\begin{array}{l} F o r p u r e r o l l i n g \\ \Rightarrow f \leq μ m g \\ \Rightarrow \frac{10 t}{3} \leq 20 \\ \Rightarrow t \leq 6 s \end{array}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET