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Class 12th Laws of Motion

Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be

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Payal Gupta

Contributor-Level 10

Let the spring is in extension state and $a_{B} > a_{A}, {\vec{a}}_{A B} = - a_{A} \hat{i} - (- a_{B} \hat{i}) = (- a_{A} + a_{B}) \hat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B . (1)

kx = Ma . (2)

$\Rightarrow a_{B} = \frac{F}{M} - a$

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6 months ago

Class 12th Ray Optics and Optical Instruments

The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors $\vec{a}, \vec{b} a n d \vec{c}$ respectively. Then choose the correct relation for these vectors.

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Payal Gupta

Contributor-Level 10

$\hat{r} - \hat{i} = 2 c o s θ \hat{n} . . . . . . (1)$

$\hat{i} . \hat{n} = - c o s θ . . . . . . (2)$

$\Rightarrow \hat{r} = \hat{i} - 2 (\hat{i} . \hat{n}) \hat{n}$

$\Rightarrow \vec{b} = \vec{a} - 2 (\vec{a} . \vec{c}) \vec{c}$

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6 months ago

Class 12th Electric Charges and Fields

An inclined plane making an angle of 30Â° with the horizontal is placed in a uniform horizontal electric field $200 \frac{N}{C}$ as shown in the figure. A body of mass 1kg and charge 5 mC is allowed to slide down from rest at a height of 1m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.

$[g = 9.8 m / s^{2}; s i n 30 ° = \frac{1}{2}; c o s 30 ° = \frac{\sqrt[]{3}}{2}]$

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Payal Gupta

Contributor-Level 10

$N = m g c o s 30 ° + q E s i n 30 °$

$a = \frac{m g s i n θ - q E c o s θ - μ N}{m} = 2.30 m / s^{2}$

$S = u t + \frac{1}{2} a t^{2}$

$\Rightarrow t = \sqrt[]{\frac{2 l}{a}} = 1.31 s e c$

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6 months ago

Class 12th Electromagnetic Waves

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

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Payal Gupta

Contributor-Level 10

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

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6 months ago

Class 12th Alternating Current

Find the peak current and resonant frequency of the following circuit (as shown in figure).

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Payal Gupta

Contributor-Level 10

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

$Z = \sqrt[]{{(120)}^{2} + {(10 - 100)}^{2}} = 150 Ω$

$ω = \frac{1}{\sqrt[]{L C}} = \frac{1}{\sqrt[]{1 0^{- 1} * 1 0^{- 4}}} = \sqrt[]{1 0^{5}}$

$? ω = 2 π f$

$\Rightarrow f = \frac{1 0^{3}}{2 π \sqrt[]{10}} = 50 H z$

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6 months ago

Class 12th Nuclei

A radioactive simple is undergoing decay. At any time t₁, its activity is A and another time t₂, the activity is $\frac{A}{5} .$ What is the average life time for the sample?

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Payal Gupta

Contributor-Level 10

$A = A_{0} e^{- λ t_{1}}$ [Radio active decay law]

$\frac{A}{5} = A_{0} e^{- λ (t_{2} - t_{1})}$

$\Rightarrow l n 5 = λ (t_{2} - t_{1})$

$\Rightarrow A v e r a g e l i f e = \frac{1}{λ} = \frac{(t_{2} - t_{1})}{l n 5}$

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6 months ago

Class 12th Mechanical Properties of Solids

The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be

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Payal Gupta

Contributor-Level 10

Stress = Y * strain

$\frac{T_{1}}{A} = Y * - \frac{(l_{1} - l)}{l} (i)$

$\frac{T_{2}}{A} = Y * \frac{(l_{2} - l)}{l} (i i)$

$\Rightarrow$ $\frac{T_{1}}{T_{2}} = \frac{l_{1} - l}{l_{2} - l}$

$\Rightarrow l = \frac{T_{1} l_{2} - T_{2} l_{2}}{T_{1} - T_{2}} = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

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6 months ago

Class 12th Mechanical Properties of Solids

The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be

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New answer posted

6 months ago

Class 12th Motion in a Plane

The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by

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Payal Gupta

Contributor-Level 10

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α * \frac{α}{2 β} - β * {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ- $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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6 months ago

Class 12th System of Particles and Rotational Motion

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be

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Payal Gupta

Contributor-Level 10

mg – T = ma. (1)

T * R = l α. (2)

a = α R. (3)

With the help of equations (1), (2) and (3), we get

$a = \frac{m g}{m + \frac{l}{R^{2}}}$

$v = \sqrt[]{2 a h} = ω R$

$\Rightarrow ω^{2} = \frac{2 m g h}{l + m R^{2}}$

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