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11 months ago

Class 11th Physics Motion in Plane

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

Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then

Reason R : Product of said heights.

Choose the correct answer :

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alok kumar singh

Contributor-Level 10

First angle, θ₁= θ, $R = \frac{u^{2} s i n 2 θ}{g}$

Another angle (θ₂ = 90 - θ) for which range will be

Same as that of θ₁ = θ

$R^{1} = \frac{u^{2} s i n 2 (90 - θ)}{g} = \frac{u^{2}}{g} s i n 2 θ = R$

at $θ_{1} = θ, h_{1} = \frac{u^{2} s i n^{2} θ}{2 g}$

$& θ_{2} = 90 - θ, h_{2} = \frac{u^{2} s i n^{2} θ_{2}}{2 g} = \frac{u^{2} c o s^{2} θ}{2 g}$

$\Rightarrow h_{1} h_{2} = \frac{1}{4^{2}} {[\frac{u^{2} s i n 2 θ}{g}]}^{2}$

$R = 4 \sqrt[]{h_{1} h_{2}}$

So, Both statement is true & Reason is correct explanation for statement 1.

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New answer posted

11 months ago

Class 11th Chemistry

According to MO theory, number of species / ions from the following having identical bond order is___________.

CN^-, NO⁺, O₂, $O_{2}^{+}, O_{2}^{2 +}$

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alok kumar singh

Contributor-Level 10

$C N^{-}, N O^{+} a n d O_{2}^{2 +}$ have bond order = 3

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11 months ago

Class 11th Chemistry

The molar heat capacity for an ideal gas at constant pressure is 20.785JK^-1 mol^-1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is __________.

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alok kumar singh

Contributor-Level 10

$C_{P, m} = C_{v, m} + R$

$C_{v, m} = 20.785 - 8.314 = 12.471 J K^{- 1} m o l^{- 1}$

$n = \frac{5000}{12.471 * 200} = \frac{25}{12.471} \approx 2$

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11 months ago

Class 11th Chemistry

Amongst the following, the number of oxide(s) which are paramagnetic in nature is Na₂O, KO₂, NO₂, N₂O, ClO₂, NO, SO₂, Cl₂O (Nearest Integer)

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alok kumar singh

Contributor-Level 10

$K O_{2}, N O_{2}, C l O_{2}, N O$ are paramagnetic.

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11 months ago

Class 11th Physics

The torque of a force $5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $τ$ . If the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + \hat{k},$ then the value of $τ$ will be

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Payal Gupta

Contributor-Level 10

$\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$

$\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$

$\vec{τ} = | \vec{r} * \vec{F} | = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 \\ 5 & 3 & - 7 \end{matrix} | \begin{matrix} = \hat{i} (- k_{1} - 3) - \hat{j} (- 14 - 5) + \hat{k} (6 - 10) & = - 17 \hat{i} + 19 \hat{j} - 4 \hat{k} \end{matrix}$

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11 months ago

Class 11th Work, Energy and Power

If momentum of a body is increased by 20%, then its kinetic energy increases by

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Payal Gupta

Contributor-Level 10

$\frac{Δ K}{K_{i}} * 100 = \frac{K_{f} - K_{i}}{K_{i}} * 100$

$= \frac{\frac{P_{f}^{2}}{2 m} - \frac{P_{i}^{2}}{2 m}}{\frac{P_{i}^{2}}{2 m}} * 100 = \frac{P_{f}^{2} - P_{i}^{2}}{P_{i}^{2}} * 100$

$= \frac{{(1.2 P_{i})}^{2} - {(P_{i})}^{2}}{P_{i}^{2}} * 100 = \frac{1.44 P_{i}^{2} - P_{i}^{2}}{P_{i}^{2}} * 100$

= 44%

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11 months ago

Class 11th Physics Laws of Motion

Two bodies of masses m₁ = 5kg and m₂ = 3kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m₁ will be: [Take g = 10 ms^-2]

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Payal Gupta

Contributor-Level 10

m₁ = 5kg

m₂ = 3kg

As m₁ is in rest

T = m_gg = 30

N = m₁ g cos

T = $m_{1} g s i n θ \Rightarrow 50 s i n θ = 30$

sin $= \frac{3}{5} \Rightarrow θ = 37 °$

$∴ N = 50 c o s θ = 50 c o s 37 ° = 50 * \frac{4}{5} = 40 N$

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11 months ago

Class 11th Physics Motion in Straight Line

A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t₂.

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Payal Gupta

Contributor-Level 10

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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11 months ago

Class 11th Physics

An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

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Payal Gupta

Contributor-Level 10

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} * 1 * 10 * 6400 k m$

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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11 months ago

Class 11th Physics

What are the two conditions for equilibrium?

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Syed Aquib Ur Rahman

Contributor-Level 10

The two conditions for equilibrium in physics are

Translational Equilibrium: The net force acting on the object must be zero, which means all the horizontal and vertical forces cancel each other out.
Rotational Equilibrium: The net torque must also be zero, indicating that the clockwise and counter-clockwise torques are balanced.

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