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Physics Laws of Motion

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2 months ago

Physics Laws of Motion 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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2 months ago

Physics Laws of Motion 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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2 months ago

Physics Laws of Motion 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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2 months ago

Physics Laws of Motion Class 11th

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force $F$ at $a$ point $b$ above its centre of mass (see figure). If the coefficient of friction is $μ = 0.4$ , the maximum possible value of $100 * \frac{b}{a}$ for box not to topple before moving is

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alok kumar singh

Contributor-Level 10

For no toppling

$F (\frac{a}{2} + b) \leq m g \frac{a}{2}$

$μ \frac{a}{2} + μ b \leq \frac{a}{2}$

$0.2 a + 0.4 b \leq 0.5 a$

$0.4 b \leq 0.3 a$

$b \leq \frac{3 a}{4}$

$b \leq 0.75 a$ (in limiting case)

But is not possible as maximum value of $b$ can be equal to $0.5 a$ only.

$∴ {(100 \frac{b}{a})}_{m a x} = 50.00$

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2 months ago

Physics Laws of Motion 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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2 months ago

Physics Laws of Motion Class 11th

A car is moving with speed of 150 km/h and after applying the break it will move 27m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling _________m distance.

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Vishal Baghel

Contributor-Level 10

Stopping distance = $\frac{v^{2}}{2 a} = d$ $\frac{^{}}{}$

If speed is made $\frac{1}{3} r d$ $\frac{}{}$

$d^{1} = \frac{d}{9}, d^{1} = \frac{27}{9} = 3 m$

Braking acceleration Remains same.

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2 months ago

Physics Laws of Motion Class 11th

A uniform chain of 6m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is………….m.

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Payal Gupta

Contributor-Level 10

F.B.D. of hanging length

F.B.D. of chain lying on the table

f = T = $μ N$

$\Rightarrow λ x g = μ λ (L - x) g$

$x = 0.5 (6 - x)$

$\Rightarrow x = 3 - 0.5 x$

$1.5 x = 3$

x = 2m

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2 months ago

Physics Laws of Motion Laws of Motion

A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2} .$ the spring constant is _________ Nm^-1. (Use g = 10 ms^-2)

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Vishal Baghel

Contributor-Level 10

COME

$k \cdot E_{i} + P \cdot E_{i} = k \cdot E_{f} + P \cdot E_{f}$

$\Rightarrow 0 + m g (h + \frac{h}{2}) = 0 + \frac{1}{2} k {(\frac{h}{2})}^{2}$ $\Rightarrow \frac{3 h}{2} m g = \frac{1}{2} k \frac{h^{2}}{4}$

$\Rightarrow 3 m g h = \frac{k}{4} h^{2} \Rightarrow 12 \frac{m g}{h} = k \Rightarrow k = \frac{12 * 0.1 * 10}{0.1}$

k = 120Nm^-1

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2 months ago

Physics Laws of Motion Class 11th

A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

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Payal Gupta

Contributor-Level 10

Is minimum at the highest position of the circular path.

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2 months ago

Physics Laws of Motion Laws of Motion

A block of mass 10 kg, starts sliding on a surface with an initial velocity of 9.8 ms^-1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is :

[used g = 9.8 ms^-2]

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Vishal Baghel

Contributor-Level 10

a µg = 0.5 * 9.8 = 4.9 m/sec²

u = 9.8 m/sec

S =?

v = 0

v² = u² + 2as

^{$\Rightarrow 0 = 9 . 8^{2} - 2 * 4.9 s$}

^{$\Rightarrow 0 = 9.8 * 9.8 - 9.8 s$}

s = $\frac{9.8 * 9.8}{9.8} = 9.8 m$

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