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Overview Class 11th

A body of mass M moving at speed V₀ collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles q₁ and q₂ with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles q₁ and q₂ will be equal, is:

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Raj Pandey

Contributor-Level 9

$M V_{0} = M V_{1} + m V_{2} . . . . (i)$

∴ M V_{1} = m V_{2}

. (ii)

$M V_{0}^{2} = M V_{1}^{2} + m V_{2}^{2}$ . (iii)

$M V_{0}^{2} = M V_{1}^{2} + m {(\frac{M V_{1}}{m})}^{2} [u s i n g e q^{n} (i i)]$

$(1 + \frac{M}{m}) = 4$

${(\frac{M}{m})}_{m a x} = 3$

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Overview Class 11th

A car is moving on a plane inclined at 30° to the horizontal with an acceleration of 10ms^-2 parallel to the upward. A bob is suspended by a string from the roof of the car. The angle in degree which the string makes with the vertical is______. (Take g = 10ms^-2)

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Raj Pandey

Contributor-Level 9

$θ = \frac{60 °}{2} = 30 °$

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New answer posted

a year ago

Overview Class 11th

The initial mass of a rocket is 1000kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20ms^-². The gases come out at a relative speed of 500ms^-¹ with respect to the rocket: [Use g = 10m/s²]

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Vishal Baghel

Contributor-Level 10

$V_{r e l} \frac{d m}{d t} - m g = F_{e x t}$

=>500 $\frac{d m}{d t} - 1000 * 10 = 1000 * 20$

$\frac{d m}{d t} = 60 k g / s$

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New answer posted

a year ago

Overview Class 11th

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $ω .$ The coefficient of static friction between the bottom of the beaker and the surface of the disc is $ω .$ The beaker will revolve with the disc if :

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Raj Pandey

Contributor-Level 9

F. B. D of Beaker

By NLM2

$\begin{array}{l} f = m a = m ω^{2} R \\ \Rightarrow m ω^{2} R \leq μ N \\ R \leq μ g / ω^{2} \end{array}$

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New answer posted

a year ago

Overview Class 11th

For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional forces must be applied and at what angle with positive x-axis so that the net acceleration of body is zero?

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Payal Gupta

Contributor-Level 10

Let additional force is $\vec{F}$ , so

$\vec{F} + 5 (+ \hat{i}) + 6 (- \hat{i}) + 7 (+ \hat{j}) + 8 (- \hat{j}) = \vec{0}$

$\Rightarrow \vec{F} - (\hat{i} + \hat{j}) = \vec{0} \Rightarrow \vec{F} = \hat{i} + \hat{j} \Rightarrow F = \sqrt[]{2} N, 45 °$

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New answer posted

a year ago

Overview Class 11th

Two billiard balls of mass 0.05 kg each moving in opposite directions with 10ms^-1 collide and rebound with the same. If the time duration of contact is t = 0.005s. Then what is the force exerted on the ball due to each other?

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Payal Gupta

Contributor-Level 10

$Δ p$ = Change in momentum of each ball

$Δ p = 2 m v = 2 * 0.05 * 10 = 1 N s$

$\Rightarrow F = \frac{Δ p}{Δ t} = \frac{2 * 0.05 * 10}{0.005} = 200 N$

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New answer posted

a year ago

Overview Class 11th

A drop of liquid of density $ρ$ is floating half immersed in a liquid of density $σ$ and surface tension 7.5 * 10^-4Ncm^-1. The radius of drop in cm will be: (g = 10ms^-2)

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Payal Gupta

Contributor-Level 10

Since liquid drop is in equilibrium, so

mg = F_B + 2πRT

$\Rightarrow \frac{4}{3} π R^{3} ρ g = \frac{2}{3} π R^{3} σ g + 2 π R T$

$\Rightarrow R = \frac{3 T}{(2 ρ - σ) g} = \frac{15}{\sqrt[]{(2 ρ - σ)}} * 1 0^{- 2} m = \frac{15}{\sqrt[]{(2 ρ - σ)}} c m$

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New answer posted

a year ago

Overview Class 11th

A balloon has mass of 10g in air. The air escapes from the balloon at a uniform rate with velocity 4.5 cm/s. If the balloon shrinks in 5s completely. Then, the average force acting on the balloon will be (in dyne).

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Vishal Baghel

Contributor-Level 10

$F t = m Δ v$

$? F ? = \frac{m Δ v}{t} = \frac{10 * 1 0^{- 3} * 4.5}{100 * 5}$ 9 * 10^-4 N.

= 9 * 10^-5 N

= 9 dyne.

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New question posted

a year ago

Overview Class 11th

In two different experiments, an object of mass 5kg moving with a speed of 25 ms^-¹ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively.

Choose the correct option out of the following:

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New answer posted

a year ago

Overview Class 11th

Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F₁ to F₂ is 1: x where x = _________.

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Vishal Baghel

Contributor-Level 10

$F_{2} = 1 c o s 45 ° + 2 c o s 45 ° = 3 c o s 45 ° = \frac{3}{\sqrt[]{2}} N$

$\begin{array}{l} F_{1} + 1 c o s 45 ° = 2 s i n 45 ° \\ \frac{F_{1}}{F_{2}} = 1 : 3 \end{array}$

x = 3

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