Class 11th

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

According to the de Broglie equation, the particles like electrons have wave-like properties. The quantum mechanical model foundation is laid by this concept and it also explains the stability of electron orbits using wave behavior.

The hydrogen atom can be accurately explained by Bohr's model. However, it does not account for shielding effects, electron-electron interactions, and the wave nature of electrons for multi-electron atoms.

Quantum numbers describe the unique position and energy of an electron in an atom. These are a set of four numbers and are important for predicting chemical behavior and understanding the distribution of electrons in orbitals.

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