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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eight

When a car is approaching the observer, the frequency of horn is 100Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be $\frac{x}{3}$ Ha where x = _________.

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Vishal Baghel

Contributor-Level 10

$f_{1} = 100 = f o (\frac{C}{C - V_{S}})$

C = Speed of sound

Vs = Speed of source

f₂ = 50 = fo = $(\frac{C}{C + V_{s}})$

$\frac{f_{1}}{f_{2}} = 2 = \frac{C + V_{S}}{C - V_{S}}$

$f o = \frac{200}{3} = \frac{x}{3}$

x = 200

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Mass numbers of two nuclei are in the ratio of 4 : 3. Their nuclear densities will be in the ratio of

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alok kumar singh

Contributor-Level 10

$R = R ? A^{1 / 3}$

$δ = \frac{m a s s}{v o l u m e} = \frac{m A}{\frac{4}{3} π R^{3}} = \frac{m A}{4 / 3 π R_{0}^{3} A}$

$\Rightarrow$ Density is independent of mass number

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In an experiment to determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young's modulus $(Δ Y)$ is found to be x * 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

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Vishal Baghel

Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 10^-3 m

$\frac{F}{A} = y \frac{Δ L}{L}$

$y = \frac{F L}{A Δ L} = \frac{(m g) * 1}{(\frac{π d^{2}}{4}) * 0.4 * 1 0^{- 3}}$

$Δ y = 0.1 * 0.199 * 1 0^{12} = 1.99 * 1 0^{10}$

= 1.99

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

Two masses M₁ and M₂are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M₂is twice that of M₁. The acceleration of the system is a₁. When the mass M₂ is thrice that of M₁ the acceleration of the system is a₂. The ratio $\frac{a_{1}}{a_{2}}$ will be:

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alok kumar singh

Contributor-Level 10

$a_{1} = \frac{m_{2} g - m_{1} g}{m_{1} + m_{2}} = \frac{2 m g - m g}{3 m} = \frac{g}{3}$

$a_{2} = \frac{3 m g - m g}{4 m} = \frac{g}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{2}{3}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Vishal Baghel

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 * 3 R}}$

$x = \frac{g}{2} = 5$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

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Vishal Baghel

Contributor-Level 10

y = x5 (1 - x) = x tanθ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y - component of initial velocity

= u sinθ

= $= \sqrt[]{26} * \frac{5}{\sqrt[]{26}}$

= 5 m/s

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with B = 12 mT. The value of E required for and electron of energy 728 eV moving along the positive x-axis to pass undeflected is:

(Given. Mass of electron = 9.1 * 1031kg)

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alok kumar singh

Contributor-Level 10

$K E = 728 * 1.6 * 1 0^{- 19} = \frac{1}{2} * 9.1 * 1 0^{- 31} * V^{2}$

V = 16 * 106 m/s

For Lorentz force to be zero

eE = eVB

$\Rightarrow E = v B = 16 * 1 0^{6} * 12 * 1 0^{- 3}$

= 192 * 103 V/m

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

Two concentric circular loops of radii r₁ = 30cm and r₂ = 50cm are placed in X – Y plane as shown in the figure. A current I = 7A is flowing through them in the direction as shown in figure. The net magnetic momentum of this system of two circular loops is approximately:

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alok kumar singh

Contributor-Level 10

$\vec{M} = I A_{1} - I A_{2}$

$\Rightarrow M = - 7 * \frac{22}{7} * [\frac{25}{100} - \frac{9}{100}] \hat{k}$

$\Rightarrow \vec{M} = \frac{- 7}{2} \hat{k} A m^{2}$

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is E₁. Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E₂. The ratio E₁/E₂ will be:

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