Class 11th

This is a multiple choice answer as classified in NCERT Exemplar

(c) 2Tsin $\theta$ -mg=0

2Tsin $\theta$ =mg

Total horizontal forces = Tcos $\theta -Tcos\theta =0$

T=mg/2sin $\theta$

As mg is constant T $\propto \frac{1}{sin\theta }$

T_max= mg/sin $\theta$ _min

Sin $\theta$ _min=0, $\theta$ _min= 0

T_min=mg/2sin $\theta$ _max

_{$sin\theta$ max}= 1, $\theta$ =90⁰

This is a multiple choice answer as classified in NCERT Exemplar

Y= $\frac{stress}{strain}=\frac{\frac{F}{A}}{\frac{?L}{L}}=\frac{F}{\pi ({D/2)}^2}*\frac{L}{?L}$

D²= $\frac{4FL}{\pi ?LY}$

D= $\sqrt[]{\frac{4FL}{\pi ?LY}}$

$\frac{D_{copper}}{D_{iron}}=\sqrt[]{\frac{Y_{iron}}{Y_{copper}}}$

This is a multiple choice answer as classified in NCERT Exemplar

(c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes. The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

This is a multiple choice answer as classified in NCERT Exemplar

(d) Y young's modulus is inversely proportional to temperature, so if we increase temperature , young's modulus decreases.

This is a multiple choice answer as classified in NCERT Exemplar

(d) breaking stress = breaking force/area of cross section

Breaking force will not depend upon length.

This is a multiple choice answer as classified in NCERT Exemplar

(b) No frictional force exists in case of ideal fluid. Hence tangential forces are zero.

This is a short answer type question as classified in NCERT Exemplar

As ivory ball is more elastic than wet clay ball. So ivory ball tries to regain its original shape quickly, hence more energy and momentum transferred to ivory than clay ball. So ivory ball rises high

This is a short answer type question as classified in NCERT Exemplar

Length of steel cable l=9.1m

Radius r= 5mm= $5*{10}^{-3}$

Tension in the cable F =800N

Young's modulus of steel Y= 2 $*\frac{{10}^{11}N}{m^2}$

So $?l=\frac{F}{\pi r^2}*\frac{l}{Y}=\frac{800}{3.14*{(5*{10}^{-3})}^{-2}}*\frac{9.1}{2*{10}^{11}}=4.64*{10}^{-4}m$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- for small objects  sizes less than 100 m centre of mass very close with the centre of gravity of the body. But when the size of object increases it weight changes and its CM and CG become far from each other.