Laws of Motion

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New answer posted•a year ago

Laws of Motion Class 11th

A body of mass 2kg moves under a force of (2i + 3j + 5k)N. It starts from rest and was at the origin initially. After 4 s, its new coordinates are (8, b, 20). The value of b is --------. (Round off the Nearest Integer)

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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New answer posted•a year ago

Laws of Motion Class 11th

Statement I: A cyclist is moving on an un-banked road with a speed of 7 kmh?¹ and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2 The cyclist will not slip and pass the curve. (g = 9.8m/s²)
Statement II: If the road is at an angle of 45°, cyclist can cross of the curve of 2m radius with the speed of 18.5 kmh?¹ without slipping
In the light of the above statements, choose the correct the answer from the options given below.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Statement-I: F? = mv²/r ≤ f? = µmg ⇒ v ≤ √ (µgR) = √ (0.2×9.8×2) = 1.98 m/s
v_cyclist = 7 km/h = 1.94 m/s. Since 1.94 m/s < 1.98 m/s, statement-I is correct.
Statement-II: v_max = √gR (tanθ+µ)/ (1-µtanθ) = .
v_min = √gR (tanθ-µ)/ (1+µtanθ) = √ (9.8×2 (tan45-0.2)/ (1+0.2tan45) = 3.65 m/s
v_cyclist = 18.5 km/h = 5.14 m/s. This is outside t

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New answer posted•a year ago

Laws of Motion Class 11th

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:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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 = MaB . (1)

kx = Ma. (2)

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

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New answer posted•a year ago

Laws of Motion Class 11th

A bob is whirled in a horizontal plane by means of a string with an initial speed of $ω rpm$ . The tension in the string is $T$ . If speed becomes $2 ω$ while keeping the same radius, the tension in the string becomes:

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$T = m \hat{?} ω^{2}$

$T^{'} = m : {(2 ω)}^{2}$

$T^{'} = 4 T$

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New answer posted•a year ago

Laws of Motion Class 11th

A $10 μ F$ capacitor is connected to a $210 V, 50 Hz$ source as shown in figure. The peak current in the circuit is nearly ( $π = 3.14$ ):

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Capacitive Reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 1 0^{- 6}}$

$= \frac{1000}{3.14}$

$V_{rms} = 210 V$

$i_{rms} = \frac{V_{rms}}{X_{C}} = \frac{210}{X_{C}}$

$Peak current = \sqrt[]{2} i_{ms} = \sqrt[]{2} \times \frac{210}{1000} \times 3.14 = 0.932$

$? 0.93 A$

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New answer posted•a year ago

Laws of Motion Class 11th

The output $(Y)$ of the given logic gate is similar to the output of an/a

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$Y_{1} = \bar{A \cdot A}$

$= \bar{A}$

$Y_{2} = \bar{B + B}$

$= \bar{B}$

$Y = \bar{Y_{1} + Y_{2}}$

$= \bar{\bar{A} + \bar{B}}$

$= \bar{\bar{A}} \cdot \bar{\bar{B}}$

$= A \cdot B is similar to output of AND Gate$

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New answer posted•a year ago

Laws of Motion Class 11th

When a body slides down from rest along a smooth inclined plane making an angle of 30° with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and though the same distance, it takes time aT, where a is a constant greater than 1. The co-efficient of friction between the body and the rough plane is $\frac{1}{\sqrt[]{x}} (\frac{α^{2} - 1}{α})$ where x =___________.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

for smooth surface

$a = g s i n 30 ° = \frac{g}{2}$

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

$S_{1} = \frac{1}{2} \frac{g}{2} t^{2} = \frac{g}{4} t^{2} . . . . . . . (i)$

for rough Surface

$S = \frac{1}{2} \frac{g}{2} (1 - μ \sqrt[]{3}) α^{2} t^{2} . . . . . . . . (i i)$

By (i) and (ii)

$μ = \frac{1}{\sqrt[]{3}} (\frac{α^{2} - 1}{α^{2}}) x = \sqrt[]{3}$

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New answer posted•a year ago

Laws of Motion Class 11th

A horizontal force 10 N is applied to a block $A$ as shown in figure. The mass of blocks $A$ and $B$ are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block $A$ on block $B$ is :

15 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$F = (M_{1} + M_{2}) a$

$a = \frac{10}{2 + 3} = 2 m s^{- 2}$

$F^{'} = M_{2} (2) = 3 \times 2 N = 6 N$

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New answer posted•a year ago

Laws of Motion Class 11th

Two bodies $A$ and $B$ of same mass undergo completely inelastic one dimensional collision. The body $A$ moves with velocity $v_{1}$ while body $B$ is at rest before collision. The velocity of the system after collision is $v_{2}$ . The ratio $v_{1} : v_{2}$ is

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Sol. Before collision $\Rightarrow$

It undergoes completely inelastic collision

Using conservation of linear momentum

Initial momentum = Final momentum

$\Rightarrow m v_{1} = m v_{2} + m v_{2}$

$\Rightarrow m v_{1} = 2 m v_{2}$

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{2}{1}$

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New answer posted•a year ago

Laws of Motion Class 11th

A sheet block of 10kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration 0.2 m/s². The normal reaction R’ by the floor if mass of the iron cylinders are equal and of 20kg each, is_________N.

[Take g = 10 m/s² and $μ_{s} = 0.2]$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

70 g – N = 70 a = 70 * 0.2 = 14

->N = 700 – 14 = 686 N

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