physics ncert solutions class 11th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-c, d

Explanation- Gravitation force F= $\frac{GMm}{r^2}$   where M is mass of sun and m is mass of earth.

When G decrease force will also decrease so according to this it will not in the fixed circular orbits around the sun. as when radius increases the force will decreases and soon earth leave the solar system.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c, d

Explanation-due to the huge amount of charges on the sun and the earth electrostatics force of attraction will be large. Gravitational force is also attractive in nature have both forces will be added and obeys kepler's law also.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c, d

Explanation- G'=10G

Consider the diagram

Force on the objects due to the earth = $\frac{G^{\text{'}}Mm}{R^2}=\frac{10GMm}{R^2}$

= (10g)m= 10mg

Force on the objects due to the sun F= $\frac{GM\text{'}m}{r^2}=\frac{GMm}{10r^2}$

As r>>R  so F will be very small

So effect of sun will be neglected

Now g'=10g

Hence weight of person =10mg

So gravity pull on the person will increase, due to it walking on ground would be more difficult

Critical velocity V_c is proportional to g

So V_{c $\propto g$}

V'>g

V_c'>V_c hence rain will fall faster. To overcome the increased gravitational force of the earth the aeroplane will have to travel much faster.

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

Bulk modulus of rubber is 9.8*10⁸ N/m²

density of sea water is 10³ kg m^-3

percentage decrease in volume $\frac{?V}{V}*100=0.1$

$\frac{?V}{V}=\frac{1}{1000}$

Let the rubber ball be taken up to depth h.

Change in pressure =h $\rho$ g

Bulk modulus K= $\frac{p}{\frac{?V}{V}}=\frac{h\rho g}{\frac{?V}{V}}$

So h = $\frac{k*?V/V}{\rho g}=\frac{9.8*{10}^8*1/1000}{{10}^3*9.8}=100m$

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

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

Coefficient of thermal expansion $\alpha$ = 10^-5

Length =1m

Area = 10^-4m²

Increase in temperature $?$ t= 200⁰C

Tension produced in steel rod F = YA $\alpha ?t$ =2 $*{10}^{11}*{10}^{-4}*{10}^{-5}*200$

= 4 $*{10}^{4}N$

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

Young's modulus Y= $\frac{F}{A}*\frac{L}{l}$

For first wire Y= $\frac{f}{\pi r^2}*\frac{L}{l}$

For second wire Y= $\frac{2f}{\pi ({2r)}^2}*\frac{2L}{l^{\text{'}}}$

From above equations $\frac{f}{\pi r^2}*\frac{L}{l}=\frac{f}{\pi r^2}*\frac{L}{l\text{'}}$

So l=l' hence Y will also same.

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

Young's modulus Y= $\frac{F}{A}*\frac{l}{?l}$

For a perfectly rigid body change in length $?l=0$

Y= $\infty$ so young's modulus for perfectly rigid body is infinite

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

Work done =1/2F $*?$ l

As spring of both are equally compressed

So Young's modulus Y= $\frac{F}{A}*\frac{l}{?l}$    as $?l$ is inversely proportional to young's modulus

And also we know W is inversely proportional to young's modulus

So $\frac{W_{steel}}{W_{copper}}=\frac{Y_{copper}}{Y_{steel}}1$

W_steelcopper 

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

stress= $\frac{magnitudeofinternalreactionforce}{areaofcrosssection}$

Therefore stress is scalar quantity not vector quantity.

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

Young's modulus Y= stress/longitudinal strain

For same longitudinal strain, $\frac{Y_{steel}}{Y_{rubber}}=\frac{{stress}_{steel}}{{stress}_{rubber}}$

Y_steel>Y_rubber

so we can say that  stress_steel > stress_rubber