Class 11th

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

Dimensions of energy is E= [ML²T^-2]

Mass m= [M]

Dimension of E= [ML²T^-2]

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

Dimensions of G= [M^-1L³T^-2]

By using these values [P]= [ML²T^{-2] $*\left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-1}\right]$ 2 $[{\mathrm{M}]}^{-5}*[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}]$ -2}

= [M^1+2-5+2L^2+4-6T^-2-2+4]

= [M⁰L⁰T⁰]

After we know that P is dimensionless quantity

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

As we know that X= a² b³ c^5/2 d^-2

Maximum percentage error in X is $\frac{?X}{X}*100=\pm \left. [2\left(\frac{?a}{a}*100\right)+3\left(\frac{?b}{b}*100\right)+\frac{5}{2}\left(\frac{?c}{C}*100\right)+2\left(\frac{?d}{d}*100\right)\right]$

= $\pm \left. [2\left(1\right)+3\left(2\right)+\frac{5}{2}\left(3\right)+2\left(4\right)\right]\%$

$\pm [2+6+\frac{15}{2}+8]$

$=\pm 23.5\%$

Mean absolute error in X= $\pm 0.24$ rounding off to significant value.

And calculated value would be 2.8 rounding off upto two digits.

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