Question

In column-I, some situations are shown and in column-II, information about their resulting motion is given. Select the correct answer using the codes given below the columns

	Column I		Column II
p.	A uniform solid sphere of mass $1kg$ and radius $1m$, ${\mu }_s=0.05$	1.	Friction will be in the direction of acceleration of centre of mass of body.
q.	A uniform body of mass $1kg$, $r=\frac{1}{2}m$, $R=1m$, $I$ (about axis passing through centre and perpendicular to the plane of paper) = $2kg{m}^2,{\mu }_s=0.3$	2.	Friction will be opposite to the direction of acceleration of centre of mass of the body.
r.	A uniform solid cylinder released on a fixed incline plane $m=2kg,R=1m,{\mu }_s=\frac{2}{5}$	3.	Body rotates clockwise
s.	A uniform body of mass $1kg,r=\frac{1}{2}m,R=1m$, $I$(about axis passing through centre and perpendicular to the plain of the paper) $=2kg{m}^2,{\mu }_s=0.5$ (String tightly wound on inner radius is pulled).	4.	Body rotates anticlockwise

• A. p-1; q- 2; r-4; s-3
• B. p-2; q- 1; r-3; s-4
• C. p-4; q- 1; r-2; s-3
• D. p-4; q- 2; r-3; s-1

Answer 1

The correct option is A p-1; q- 2; r-4; s-3

p.
Let friction be static
$F+f=maFR-fR=\frac{2}{5}m{R}^2\alpha \Rightarrow F-f=\frac{2}{5}mR\alpha =\frac{2}{5}ma(a=R\alpha )\frac{F+f}{F-f}=\frac{5}{2}\Rightarrow f=\frac{3F}{7}=\frac{6}{7}NBut{f}_L=\mu mg=0.5N$
Therefore, friction is kinetic

q.
$4-f=ma=a4\times \frac{1}{2}+f\times 1=2af=2N{f}_L=\mu mg=3N$

r. $a=\frac{g\sin \theta }{1+\frac{I}{m{R}^2}}=\frac{10\times \frac{1}{2}}{1+\frac{1}{2}}=\frac{10}{3}m/s^{2}mg\sin \theta -f=maf=\frac{mg}{2}-ma=10-\frac{20}{3}=\frac{10}{3}N{f}_L={\mu }_{s}N={\mu }_{s}mg\sin \theta =\frac{2}{5}\times 2\times 10\times \frac{\sqrt{3}}{2}=4\sqrt{3}N$
Therefore, friction is static

s.
$f_L=5Nf=\frac{10}{3}N\Rightarrow StaticfrictionF-f=ma\Rightarrow 4-f=1af\times 1-4\times \frac{1}{2}=2\times \frac{a}{R}\Rightarrow f-2=2a\Rightarrow a=\frac{2}{3}m/s^2$