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3 months ago

Class 11th Physics Motion in Plane

The ranges and heights for two projectiles projected with the same initial velocity at angles 42° and 48° with the horizontal are R₁, R₂ and H₁, H₂ respectively. Choose the correct option:

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alok kumar singh

Contributor-Level 10

Range R = $\frac{u^{2} s i n 2 θ}{g}$

For 42° and 48° Range will be same

$H_{m a x} α$ maximum q

So maximum height will be for 48°

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3 months ago

Class 11th Work, Energy and Power

A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8 $\sqrt[]{g h} .$ The value of work done by the air-friction is:

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alok kumar singh

Contributor-Level 10

Work done is equal to change in K.E.

$S o W_{1} + W_{2} = \frac{1}{2} M {(0.8 \sqrt[]{g h})}^{2} - 0$ W₁^-> work done by mg

$m g h + W_{2} = \frac{1}{2} m * 0.64 g h$ W₂ -> work done by air friction

$W_{2} = 0.32 m g h - m g h = - 0.68 m g h$

W₂ = -0.68 mgh

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3 months ago

Class 11th Oscillations

A bob of mass 'm' suspended by a thread of length I undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by 1/3^rd of the original length, then the time period of the simple harmonic oscillations will be:-

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Vishal Baghel

Contributor-Level 10

Time period T = 2π $\sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T_{i} = T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow T_{f} = T' = 2 π \sqrt[]{\frac{4 l / 3}{g (1 - \frac{ρ}{σ})}} = 2 π \sqrt[]{\frac{16 l}{9 g}}$

$\Rightarrow T' = \frac{4}{3} T$

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Class 11th Physics Laws of Motion

Statement-I: Two forces $(\vec{P} + \vec{Q}) a n d (\vec{P} - \vec{Q}) w h e r e \vec{P} ⊥ \vec{Q},$ when act at an angle θ₁ to each other the magnitude of their resultant is $\sqrt[]{3 (P^{2} + Q^{2})},$ when they act an angle θ₂, the magnitude of their resultant becomes $\sqrt[]{2 (P^{2} + Q^{2})} .$ This is possible only when θ₁ < ₂.

Statement-II: In the situation given above.

θ₁ = 60° and θ₂ = 90°

In the light of the above statements, choose the most appropriate answer from the options given below:-

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Vishal Baghel

Contributor-Level 10

$\vec{x} = | \vec{P} + \vec{Q} | = \sqrt[]{P^{2} + Q^{2}}$

=> $\vec{y} = | \vec{P} - \vec{Q} | = \sqrt[]{P^{2} + Q^{2}}$

$| \vec{x} + \vec{y} | = \sqrt[]{x^{2} + y^{2} + 2 x y}$

$\Rightarrow \sqrt[]{3 (P^{2} + Q^{2})} = \sqrt[]{{(\sqrt[]{P^{2} + Q^{2}})}^{2} + {(\sqrt[]{P^{2} + Q^{2}})}^{2} + 2 (P^{2} + Q^{2}) c o s θ_{1}}$

$\Rightarrow θ_{1} = 60 °$

=>Using same formula : θ₂ = 90°

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3 months ago

Class 11th NCERT Class 11 Maths

The line 12x cosq + 5y sinq = 60 is tangent to which of the following curves?

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Raj Pandey

Contributor-Level 9

Since the line $\frac{x}{5} c o s θ + \frac{y}{12} s i n θ = 1$ is the equation of tangent to the ellipse $\frac{x^{2}}{5^{2}} + \frac{y^{2}}{1 2^{2}} = 1$ at (5 cos q, 12 sin q). Hence option (A) is correct option.

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3 months ago

Class 11th NCERT Class 11 Maths

Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new number are in an arithmetic progression with common difference d. If the fourth term of GP is 3r², then r² – d is equal to:

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Raj Pandey

Contributor-Level 9

Let $? \frac{a}{r}$ , a, ar are in G.P. $∴ \frac{a}{r},$ 2a, ar are in A.P.

$2 a - \frac{a}{r} = a r - 2 a \Rightarrow 2 - \frac{1}{r} = r - 2 \Rightarrow r^{2} - 4 r + 1 = 0 \Rightarrow r = 2 + \sqrt[]{3}$

$A / q, a r^{2} = 3 r^{2} \Rightarrow a = 3$

$\begin{array}{l} ∴ d = a r - 2 a = 3 \sqrt[]{3} \\ ∴ r^{2} - d = 7 + \sqrt[]{3} \end{array}$

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3 months ago

Class 11th Laws of Motion

A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is: Given m = 8 kg, M = 16kg. Assume all the surfaces shown in the figure to be frictionless.

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alok kumar singh

Contributor-Level 10

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9} g$

Now for 8 kg,

$8 g s i n 30 ° + 8 a c o s 30 ° = m a_{1}$

$a_{1} = g * \frac{1}{2} + \frac{\sqrt[]{3}}{9} g * \frac{\sqrt[]{3}}{2}$

$= \frac{2}{3} g$

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3 months ago

Class 11th Work, Energy and Power

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is:-

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Vishal Baghel

Contributor-Level 10

Using conservation of linear momentum:

=> 40 * 3m = 60 * m + v * 2m

=> v = 30 m/s

$K E_{i} = \frac{1}{2} * 3 m * {(40)}^{2}$

$K E_{f} = \frac{1}{2} * m * {(60)}^{2} + \frac{1}{2} * 2 m * {(30)}^{2}$

$\Rightarrow \frac{K E_{f}}{K E_{i}} = \frac{54}{48}$

Fractional change in kinetic energy = 1 $- \frac{K E_{f}}{K E_{i}} = \frac{1}{8}$

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3 months ago

Class 11th NCERT Class 11 Maths

Thew length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (>R) respectively from the origin, is:

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Raj Pandey

Contributor-Level 9

Length of latus rectum = 4 (distance between vertex and focus) = 4 (S – R)

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3 months ago

Class 11th Physics

Statement-I: If three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ are represented by three sides of triangle and, ${\vec{F}}_{1} + {\vec{F}}_{2} = - {\vec{F}}_{3},$ then these three forces are concurrent forces and satisfy the condition for equilibrium.

Statement-II: A triangle made up of three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ as its sides taken in the same order, satisfy the condition for translator equilibrium.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Vishal Baghel

Contributor-Level 10

For equilibrium net force acting on the system should be zero.

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