Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation - we know that $\theta$ =15⁰ and R= 50m

Range, R= $\frac{u^{2}sin2\theta }{g}$

50 = $\frac{u^{2}sin2*15}{g}$

So u²= 980

So u = $\sqrt[]{980}=7*2*\sqrt[]{5}$ m/s

So u= 14 $*2.23$ = 31.304m/s

$\theta$ = 45⁰, R= $\frac{u^{2}sin2\left(45\right)}{9.8}=\frac{u^2}{g}$

R= $\frac{({14\sqrt[]{5)}}^2}{g}$ = $\frac{14*14*5}{9.8}$ = $100m$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation – ler r makes an angle $?$ with positive x axis component if r along x-axis.

So  r_x= |r|cos $?$

(r_x)_maximum= |r| (cos $?$ )_maximum

= |r|cos $0$ = r

$?$ =0⁰

R is along positive x-axis.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b 

Explanation – u= a? + b? as u in the first quadrant, hence both components a and b will be positive. For v= p? + q? , as it is positive x direction and located downward hence x component p will be positive and y component q will be negative.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation- a scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation-A = i+j

 B = i-j

 A.B=|A|B|cos $\theta$

(? +? ). (? -? ) = $\left(\sqrt[]{1+1}\right)\left(\sqrt[]{1+1}\right)$ cos $\theta$

Where $\theta$  is the angle between A and B

Cos $\theta$ = $\frac{1-0+0-1}{\sqrt[]{2\sqrt[]{2}}}$ = $\frac{0}{2}=0$

$\theta$  = 90^o

This is a multiple choice answer as classified in NCERT Exemplar

When the gas is compressed adiabatically, the total work done on the gas increases its internal energy which in turn increases the KE of the gas molecule and hence, the collisions between molecules also increases.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- - given |A|=2 and |B|=4

a) |A $*B$ |=AB cos $\theta$ = 0

so 2 $*4$ cos $\theta$ =0

so $\theta =90$  so it matches with ii

b) |A $*$ B|= ABcos $\theta$ =8

2 $*4$ cos $\theta$ =8

So $\theta$ = 0 so it matches with option i

c) |A $*$ B|= ABcos $\theta$ =4

so $\theta$ =60 so it matches with option iv

d) |A $*$ B|= ABcos $\theta$ =-8

so $\theta$ =180 so it matches with option iii

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- given |A|=2 and |B|=4

a)|A $*B$ |=AB sin $\theta$ = 0

so 2 $*4$ sin $\theta$ =0

so $\theta =0$ so it matches with iv

B) |A $*$ B|= ABsin $\theta$ =8

2 $*4$ sin $\theta$ =8

So $\theta$ = 90 so it matches with option iii

c) |A $*$ B|= ABsin $\theta$ =4

so $\theta$ =30 so it matches with option i

d) |A $*$ B|= ABsin $\theta$ =4 $\sqrt[]{2}$

so $\theta$ =45 so it matches with option ii

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) pV=nRT

(a) when pressure p =constant

volume is directly proportional to temperature

(b) when T= constant

from PV =constant

so the graph is rectangular hyperbola

(c) when V =constant

from pressure is directly proportional to temperature.

So the graph is straight the passes through the origin.

PV $\propto T$

PV/T=constant

So the graph hence through origin. So d is correct.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – here A and B vectors are joint by head and tail. So C= A+B

(a) from fig iv it is clear that c=a+b

(a) From fig iii it is clear that c+b=a so a-c=b

(b) From fig I it is clear that b=a+c so b-a =c

(c) From ii it is clear that -c= a+b so a+b+c=0