ncert solutions physics class 11th

4.32

(a) Let $\mathit{\theta }\left(\mathit{t}\right)=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{{\mathit{v}}_{\mathit{o}\mathit{y}}-\mathit{}\mathit{g}\mathit{t}}{{\mathit{v}}_{\mathit{o}\mathit{x}}}\right)$ be the angle at which the projectile is fired w.r.t. the x-axis, since ${\mathit{\theta }}_0=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{4{\mathit{h}}_{\mathit{m}}}{\mathit{R}}\right)$ depends on t

Therefore tan $\theta$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $\theta$ = $\theta \left(t\right)=\frac{Vx}{Vy}=\frac{Voy-gt}{Vox}$ sin2 $\theta \left(t\right)={\tan }^{-1}?\left(\frac{Voy-gt}{Vox}\right)$ /2g…….(1)

R = $h_{max}$ sin2 $u^2$ /g …….(2)

Dividing (1) by (2)

( $\theta$ /R) = [ $u^2$ sin2 $\theta$ /2g]/[ $h_{max}$ sin2 $u^2$ /g] = $\theta$ / 4

4.31

Speed of the cyclist = 27 km/h = 7.5 m/s

Radius of the road = 80m

The net acceleration is due to braking and the centripetal acceleration

Acceleration due to braking = 0.5 m/s²

Centripetal acceleration a = v²r = (7.5)2/80= 0.703 m/s²

The resultant acceleration is given by a = sqrt ( $a_t^2$ + $a_c^2$ ) = sqrt ( ${0.5}^2$ + ${0.7}^2$ ) = 0.86 m/s²

tan $\beta$ = AC / at = 0.7/0.5,  $\beta$ = 54.5 $°$

4.30

Speed of the fighter plane = 720 km/h = 200 m/s

The altitude of the plane = 1.5 km =1500 m

Velocity of the shell = 600 m/s

$\sin ?\theta$ = 200/600

$\therefore \theta =19.47°$

Let H be the minimum altitude for the plane to fly, without being hit

Using equation H = $u^2$ sin2 (90- $\theta )/2g$

= (6002cos2 $\theta$ )/2g

= 360000 { ( 1 + cos2 $\theta$ )/2}/2 $*$ 10

= 16000 m = 16 km

4.29 The angle of projectile $\theta$ = 30 $°$

The bullet hits the ground at a distance of 3 km, Range R = 3 km

We know horizontal range for a projectile motion, R = $u^2$ sin2 $\theta$ / g

3 = $u^2$ sin60 $°$ / g

$\frac{u^2}{g}$ = 3/ sin60 $°$ = 3.464 …………… (1)

To hit a target at 5 km,

Max Range,  $R_{max}$ = $\frac{u^2}{g}$ …………. (2)

On comparing equation (1) and (2), we get

$R_{max}$ = 3.464

Hence, the bullet will not hit the target even by adjusting its angle of projection.

4.28

(a) We cannot associate the length of a wire bent into a loop with a vector.

(b) We can associate a plane area with a vector.

(c) We cannot associate the volume of a sphere with a vector.

4.27 No in the both the cases.

A physical quantity which is having both direction and magnitude is not necessarily a vector. For instance, in spite of having direction and magnitude, current is a scalar quantity. The basic necessity for a physical quantity to fall in a vector category is that it ought to follow the “law of vector addition”.

As the rotation of a body about an axis does not follow the basic necessity to be a vector i.e. it does not follow the “law of vector addition”.

4.26 Yes and No

A vector in space has no distinct location. The reason behind this is that a vector stays unchanged when it displaces in a way that its direction and magnitude do not change.

A vector changes with time. For instance, the velocity vector of a ball moving with a specific speed fluctuates with time.

Two equivalent vectors situated at different locations in space do not generate the same physical effect. For instance, two equivalent forces acting at different points on a body to rotate the body, but the combination will not generate the equivalent turning effect.