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 Short Answer Type Question 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 Question 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

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

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

Explanation – a) radius of earth =6400km= 6.4 $*{10}^{6}m$

Time period = 1 day = 24 $*60*60$ = 86400s

Centripetal acceleration a= w²r= R(2 $\pi /T$ )²=4 $\pi$ ²R/T

= $\frac{4*{\left(\frac{22}{7}\right)}^2*6.4*{10}^6}{({24*60*60)}^2}$ = 0.034m/s²

b) time = 1yr=365 $*24*60*60$ days= 365=3.15 $*{10}^{7}s$

centripetal acceleration = Rw²= $\frac{4{\pi }^{2}R}{T^2}$

^{$=\frac{4*{\frac{22}{7}}^2*1.5*{10}^{11}}{({3.15*{10}^7)}^2}=5.97*{10}^{-3}m/s$ 2}

$\frac{a_c}{g}=\frac{5.97*{10}^{-3}}{9.8}=\frac{1}{1642}$

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

Explanation- when it be at position P, drops a bomb to hit a target T

Let $\theta$

Speed of the plane =720km/h = 720 $*5/18$ = 200m/s

Altitude of the plane P'T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP'=u $*t$ =200t

Vertical distance travelled by the bomb P'T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}}=17.49s$

PP'=200 (17.49)m=

tan $\theta$ =P'T/P'P=1500/200 (17.49)=0.49287= tan23⁰12'