Physics

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

Rate of flow is equal to V= $\frac{\pi }{8}\frac{p{r}^4}{\eta l}$

Dimensions of V or LHS= volume/time=L³/T= [L³T^-1]

Dimensions of P= [ML^-1T^-2]

Dimensions of  $\eta$ = [ML^-1T^-1]

Dimensions of L= [L]

Dimensions of r= [L]

Dimensions of RHS= $\frac{\left[M{L}^{-1}{T}^{-2}\right]}{\left[M{L}^{-1}{T}^{-1}\right]}\frac{\left[L^4\right]}{\left[L\right]}=\left[L^3{T}^{-1}\right]$

So they are in equal in dimensions.

So equation is correct dimensionally.

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

Energy E= [ML²T^-2]

Let M₁, L₁ and T₁ and M₂, L₂ and T₂ are fundamental quantities for two units

M₁=1kg and L₁=1m and T₁=1s

M₂= α kg, L₂= β m and T₂= γ s

And n₁u₁=n₂u₂

This is a multiple choice answer as classified in NCERT Exemplar

Time period of simple pendulum T=2s

For simple pendulum T= $2\pi \sqrt[]{\frac{l}{g}}$  where l is length and g = acceleration due to gravity.

T_e=2 $\pi \sqrt[]{\frac{l_e}{g_e}}$

On the surface of the moon T_m= 2 $\pi \sqrt[]{\frac{l_m}{g_m}}$

$\frac{T_e}{T_m}$ = $\frac{2\pi }{2\pi }\sqrt[]{\frac{l_e}{g_e}}*\sqrt[]{\frac{g_m}{l_m}}$

T_e=T_m to maintain the second's pendulum time period

1= $\sqrt[]{\frac{l_e}{g_e}}*\sqrt[]{\frac{g_m}{l_m}}$ …………….1

But the acceleration due to gravity at moon is 1/6 of the acceleration due to gravity at earth,

g_m= $\frac{g_e}{6}$

squaring equation 1 and putting this value

1= $\frac{l_e}{l_m}*\frac{g_e/6}{g_e}=\frac{l_e}{l_m}*\frac{1}{6}$

l_m=1/6l_e = 1/6 m

This is a multiple choice answer as classified in NCERT Exemplar

Potential energy of a simple harmonic oscillator is = ½ kx²=1/2mw²x²

K=mw²

When x=0 PE=0

When x= $?A$ , PE=maximum

=1/2 mw²A²

KE of a simple harmonic oscillator =1/2 mv²

= 1/2 m [w $\sqrt[]{A^2-x^2}$ ] ²

= ½ mw² (A²-x²)

This is also parabola if plot KE against displacement x

KE= 0 at x= $?A$

KE=1/2mw²A² at x=0

Now total energy of the simple harmonic oscillator =PE+KE

= ½ mw²x²+1/2mw² (A²-x²)

TE= ½ mw²A²

So the curve according to that is

This is a multiple choice answer as classified in NCERT Exemplar

As we know x= acoswt

V =dx/dt= a (-sinwt)w=-wasinwt

V=-wasinwt

= wacos ( $\frac{\pi }{2}+wt$ )

Phase of velocity = $\frac{\pi }{2}+wt$

So difference in phse of velocity to that of phase of displacement = $\frac{\pi }{2}+wt-wt$ = $\frac{\pi }{2}$

This is a multiple choice answer as classified in NCERT Exemplar

As the particle on reference circle moves in anticlockwise direction. The projection will move from P to O towards left.

Hence in the position shown the velocity is directed from P' to P'' i.e from right to left . hence sign is negative.

This is a multiple choice answer as classified in NCERT Exemplar

In the diagram

the motion of a particle  executing SHM between A and B

Total distance travelled while it goes from A to B and returns to A is=AO+OB+BO+OA

= A+A+A+A=4A

So ratio of distance and amplitude =4A/A=4

This is a multiple choice answer as classified in NCERT Exemplar

As we know equation of SHM is x= Asinwt

V= dx/dt=Awsinwt

V_max=Awcoswt_max

= Aw

A=dv/dt=-wAwsinwt

= -w²Asinwt

A_max=-w²A

From above equations

$\frac{V_{max}}{A_{max}}$ =wA/w²A=1/w

This is a multiple choice answer as classified in NCERT Exemplar

The bob is displaced through some angle

The restoring force $\tau =-mgsin\theta$ if $sin\theta$ is small then it is $\theta$ only.

$\tau \approx -mg\theta$

So torque is directly proportional to angle.

So it clear from the above equation that its period will be harmonic

This is a multiple choice answer as classified in NCERT Exemplar

Acceleration is directly proportional to displacement.

The direction of acceleration is always towards the mean position that is opposite to displacement.