physics ncert solutions 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

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.