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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ in the $x$ -direction. The magnitude of gravitational potential on the $x$ -axis at a distance $x$ , taking its value to be zero at infinity is:

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alok kumar singh

Contributor-Level 10

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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6 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

A air bubble of radius 1 cm in water has an upward acceleration 9.8 cm^-2 . The density of water is 1 gmcm^-3 and water offers negligible drag force on the bubble. The mass of the bubble is: (g=980 cm/s²)

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alok kumar singh

Contributor-Level 10

$ρ v g - m g = m a$

$\Rightarrow \frac{ρ v g}{m} = g + a$

$\Rightarrow m = \frac{ρ v g}{g + a}$

$\Rightarrow \frac{10^{3} (\frac{4}{3} π * 10^{- 6}) (9.8)}{9.898}$

$\Rightarrow 4.15 g m$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

For a transverse wave traveling along a straight line, the distance between two peaks (crests) is 5
, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in $A \to I V; B \to I$ ) of the waves are:

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alok kumar singh

Contributor-Level 10

(n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

A small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06 T$ experiences a torque of $0.018 N m$ . The minimum work required to rotate it from its stable to unstable equilibrium position is:

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alok kumar singh

Contributor-Level 10

$\overset{?}{τ} = \overset{?}{μ} * \overset{?}{B}$

$\begin{array}{l} \Rightarrow & 0.018 = μ (0.06) (s i n ? 30^{?}) \\ \Rightarrow & μ = 0.6 \\ \Rightarrow & Work = U_{f} - U_{i} \\ \Rightarrow & 2 μ B \\ \Rightarrow & 7.2 * 10^{- 2} J . \end{array}$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3 {W m}^{- 2}$ falls normally on a polarizer (cross sectional area $3 * 10^{- 4} m^{2}$ ) which rotates about its axis with an angular speed of $31.4 r a d / s$ . The energy of light passing through the polariser per revolution, is close to:

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alok kumar singh

Contributor-Level 10

$E = (I) (t) (A) ?{c o s}^{2} ? θ?$
$\Rightarrow (3.3) [\frac{2 π}{31.4}] [3 * 10^{- 4}] * \frac{1}{2}$

$\Rightarrow 0.99 * 10^{- 4}$

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Let the mirror image of a circle c₁ : x² + y² – 2x – 6y + a = 0 in line y = x + 1 be c₂ : 5x² + 5y² + 10gx + 10fy + 38 = 0. If r is the radius of circle c₂, then a + 6r² is equal to………….

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Vishal Baghel

Contributor-Level 10

${(x - 1)}^{2} + {(y - 3)}^{2} = 10 - α$

$\begin{array}{l} x - y = - 1 \\ x + y = a + b \end{array}} \to m (\frac{a + b - 1}{2}, \frac{a + b + 1}{2})$

A' = 2m – A = (b – 1, a + 1)

$C_{2} : x^{2} + y^{2} + 2 g x + 2 f y + \frac{38}{5} = 0$

r $\sqrt[]{g^{2} + f^{2} - c}$

$= \sqrt[]{4 + 4 - \frac{38}{5}} = \sqrt[]{\frac{2}{5}}$

$C_{1} : \sqrt[]{\frac{2}{5}} = \sqrt[]{1 + 9 - α} = \sqrt[]{10 - α}$

$α + 6 r^{2} = \frac{48}{5} + 6 (\frac{2}{5}) = \frac{60}{5} = 12$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A telegraph line of length 100km has a capacity of $0.01 μ F / k m$ and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that need to be introduced in series is _________mH. (if π = $\sqrt[]{10}$ )

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Payal Gupta

Contributor-Level 10

C = 10^-6 f

f = 0.5 * 10³ = 500

$z = \sqrt[]{x_{R}^{2} + {(x_{L} - x_{C})}^{2}}$

for minimum impedance :

x_L = x_C

$\Rightarrow ω^{2} = \frac{1}{L C}$

$\Rightarrow L = \frac{1}{π^{2}} = \frac{1}{10} * 1000 = 100 m H$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

The specific heat of water $= 4200 J {k g}^{- 1} K^{- 1}$ and the latent heat of ice $= 3.4 * 10^{5} J {k g}^{- 1}$ . 100 grams of ice at $0^{\circ} C$ is placed in $200 g$ of water at $25^{\circ} C$ . The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in grams):

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alok kumar singh

Contributor-Level 10

$m (L) = m_{1} S_{1} (Δ T)$

$\Rightarrow m (3.4 * 10^{5}) = (200) (4200) (25)$

$\Rightarrow m = 61.7$

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let S = {4, 6, 9} and T = {9, 10, 11,…., 1000}.

If A = ${a_{1} + a_{2} + . . . + a_{k} : k \in N, a_{1}, a_{2}, a_{3}, . . . . ., a_{k} \in S},$ then the sum of all the elements in the set T – A is equal to…………….

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Vishal Baghel

Contributor-Level 10

T = {9, 10, 11, 12, …., 1000}

S = {4, 6, 9}

A = ${a_{1} + a_{2} + . . . . + a_{k} : k \in N, a_{i} \in S}$

Let 4 appear x no. of times

6 appear y no. of times

9 appear z no. of times

10 then set A has n element 4x + 6y + 9z

Concept : Let a and b are co-prime numbers, then members from (a - ) (b – 1) and more can be expressed in the form ax + by where x, y $\in$ {0, 1, 2, …}

So, all the number of the form

2y + 3z are (2 1). (3 – 1), ….

i.e., 6, 7, 8, 9, 10, 11, ….

form : 2 + t, t = 0, 1, 2, 3, ….

So, 6y + 9z = 3 (2y + 3z) = 3 (2 + t) = 6 + 3t, t = 0, 1, 2, 3, …

sum of element of T – A is 11

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