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11 months ago

Class 12th Motion in a Plane

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerated with $\frac{g}{6}$ vertically upwards then the time period will be:

(Where g = acceleration due to gravity)

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Payal Gupta

Contributor-Level 10

$g_{e f f} = g + a$

$= g + \frac{g}{6}$

$= \frac{7 g}{6}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$T' = \sqrt[]{\frac{6}{7}} T$

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

The dimension of mutual inductance is:

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Vishal Baghel

Contributor-Level 10

$e = - M \frac{d i}{d t}$

$M = \frac{e}{d i / d t} = M L^{2} T^{- 2} A^{- 2}$

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Class 12th Ray Optics and Optical Instruments

The efficiency of a Carnot's engine, working between stem point and ice point, will be

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Payal Gupta

Contributor-Level 10

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$η = (1 - \frac{T_{2}}{T_{1}}) * 100 = (1 - \frac{273}{373}) * 100 = 26.81 %$

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Class 12th CBSE 12th

Will there be admission without entrance after class 12th?

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

Let b₁ b₂ b₃ b₄be a 4-element permutation with $b_{i} \in {1, 2, 3, . . . ., 100} f o r 1 \leq i \leq 4 a n d b_{i} \neq b_{j}$ for $i \neq j,$ such that either b_1, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁ b₂b₃ b₄ is equal to……………

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alok kumar singh

Contributor-Level 10

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

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Class 12th Semiconductor Electronics

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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Payal Gupta

Contributor-Level 10

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

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Class 12th Semiconductor Electronics

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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

Let $P_{1} : \vec{r} . (2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let P₂be another plane which passes through the points $(2, - 3, 2), (2, - 2, - 3) a n d (1, - 4, 2)$ . If the direction ratios of the line of intersection of P₁ and P₂ be 16, , then the value of + is equal to……………

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alok kumar singh

Contributor-Level 10

2x + y – 3z = 4

$π_{2} = | \begin{matrix} x - 2 & y + 3 & z - 2 \\ 0 & 1 & - 5 \\ - 1 & - 1 & 0 \end{matrix} | = 0$

$π_{2} : - 5 x + 5 y + z + 23 = 0$

now line lying in both the planes have DR.

$\frac{a}{1 + 15} = \frac{b}{15 - 2} = \frac{c}{10 + 5}$

$\frac{a}{16} = \frac{b}{13} = \frac{c}{15}$

So direction ratio's a : b : c = 16 : 13 : 15

$x + f' (x) = f' (0)$

$f' (0) = 1 - c^{2} f r o m e q u a t i o n (i)$

$x + f' (x) = 1 - c^{2}$

$\frac{x^{2}}{2} + f (x) = (1 - c^{2}) x + d$

f(0) = 0

$f (x) = \frac{- x^{2}}{α} + (1 - c^{2}) x$

c = 3/2

$f " (x) = - 1 = 2 (c + 1)$

$f (x) = \frac{- x^{2}}{2} + (\frac{- 5}{4}) x$

now $(f (1) + f (2) + . . . . . . . f (20))$

$= \frac{- 1}{2} (1^{2} + 2^{2} + . . . . . . + 2 0^{2}) - \frac{5}{4} (1 + 2 + . . . . + 20)$

$= \frac{20 \cdot 21}{2} \cdot \frac{(- 82 - 15)}{12}$

now $2 | f (1) + f (2) + . . . . + f (20) |$

$= \frac{20 \cdot 21 \cdot 97}{12} = 3395$

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

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > 0, b > 0,$ be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (2 \sqrt[]{2} + \sqrt[]{14}) .$ If the eccentricity H is $\frac{\sqrt[]{11}}{2},$ then the value of a² + b² is equal to………….

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alok kumar singh

Contributor-Level 10

Given 2a + 2b = $4 (2 \sqrt[]{2} + \sqrt[]{14})$

$= 4 \sqrt[]{2} (2 + \sqrt[]{7})$

$b^{2} = a^{2} (\frac{11}{4} - 1)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2} a$

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11 months ago

Class 12th Laws of Motion

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

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Payal Gupta

Contributor-Level 10

By conservation of Angular momentum

L_i = L_f

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

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