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6 months ago

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

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be

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Payal Gupta

Contributor-Level 10

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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

Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to:

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alok kumar singh

Contributor-Level 10

$f (x) = \int_{1}^{3} ? \frac{\sqrt[]{x} d x}{(1 + x)^{2}} = \int_{1}^{\sqrt[]{3}} ? \frac{t . 2 t d t}{{(1 + t^{2})}^{2}}$ (put $\sqrt[]{x} = t$ )

$= {(- \frac{t}{1 + t^{2}})}_{1}^{\sqrt[]{3}} + {({t a n}^{- 1} ? t)}_{1}^{\sqrt[]{3}}$ [Applying by parts]

$\begin{array}{r} = - (\frac{\sqrt[]{3}}{4} - \frac{1}{2}) + \frac{π}{3} - \frac{π}{4} \\ = \frac{1}{2} - \frac{\sqrt[]{3}}{4} + \frac{π}{12} \end{array}$

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Sixteen

Which of the following compounds is an example of hypnotic drug?

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Vishal Baghel

Contributor-Level 10

Drug used in the treatment of sleeping problems

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6 months ago

Class 12th Gravitation

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

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Payal Gupta

Contributor-Level 10

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4$ $r$ $B$ $3$

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6 months ago

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

Match List – I with List – II

	List – I		List – II
A.	Moment of inertia of solid sphere of radius R about and tangent.	I.	5/3MR²
B.	Moment of inertia of hollow sphere of radius (R) about any tanget	II.	7/5MR²
C.	Moment of inertia of circular ring of radius (R) about its diameter.	III.	1/4MR²
D.	Moment of inertial of circular disc of radius (R) about any diameter.	IV.	1/2MR²

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Use formula for M.I.

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6 months ago

Class 12th Motion in a Plane

Motion of a particle in x – y plane is described by a set of following equations x = 4sin $(\frac{π}{2} - ω t)$ m and y = 4sin $(ω t) m .$ The path of a particle will be

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Payal Gupta

Contributor-Level 10

$x = 4 s i n (\frac{π}{2} - ω t)$ - (i)

$y = 4 s i n (ω t)$ - (ii)

From (i) and (ii) cos2 $ω t + s i n^{2} ω t = {(\frac{x}{4})}^{2} + {(\frac{y}{4})}^{2} = 1$

$\Rightarrow x^{2} + y^{2} = {(4)}^{2}$

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6 months ago

Class 12th Laws of Motion

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k² rt², where k is a constant. The power delivered to the particle by the force acting on it is given as

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Payal Gupta

Contributor-Level 10

a = k²rt²

$\Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$

$\Rightarrow v = k r t$

$a_{t} = \frac{d v}{d t} = k r$

⇒ tangential force, F_t = ma_t = mkr

$∴ P o w e r d e l i v e r e d, P = F_{t} v = (m k r) (k r t) = m k^{2} r^{2} t$

Note → Power delivered by centripetal force will be zero.

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6 months ago

Class 12th Mechanical Properties of Fluids

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.

Reason R : Coefficient of viscosity = $\frac{F o r c e}{V e l o c i t y g r a d i e n t}$

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Pressure * time $= \frac{F t}{A} = [M L T^{- 2}] [T] [L^{- 2}] = [M L^{- 1} T^{- 1}]$

Coefficient of viscosity, $η = [M L^{- 1} T^{- 1}]$

$F_{v i s c o u s} = - η A \frac{Δ v}{Δ y}$

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6 months ago

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

Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P (1, β), β > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is:

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alok kumar singh

Contributor-Level 10

$[x]^{2} + 2 [x + 2] - 7 = 0$

$\begin{array}{r} \Rightarrow [x]^{2} + 2 [x] + 4 - 7 = 0 \\ \Rightarrow [x] = 1, - 3 \\ \Rightarrow x \in [1,2) \cup [- 3, - 2) \end{array}$

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6 months ago

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

The angle of elevation of a cloud $C$ from a point $P, 200 m$ above a still take is $30^{\circ}$ . If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{\circ}$ , then $P C$ (in $m$ ) is equal to

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alok kumar singh

Contributor-Level 10

$\sum_{r = 0}^{20} ?^{50 - r} C_{6} =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . +^{30} C_{6}$

$\begin{array}{r} =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . + (^{30} C_{6} +^{30} C_{7}) -^{30} C_{7} \\ =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . + (^{31} C_{6} +^{31} C_{7}) -^{30} C_{7} \\ =^{50} C_{6} +^{50} C_{7} -^{30} C_{7} \\ =^{51} C_{7} -^{30} C_{7} \end{array}$

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