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Physics Class 11th

A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

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alok kumar singh

Contributor-Level 10

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

->5 * 0.39 * 500 = m_ice * 335

$m_{i c e} = \frac{5 * 0.39 * 500}{335}$

= 2.91 kg

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New answer posted

a year ago

Physics Physics Thermal Properties of Matter

A solid metallic cube having total surface area 24 m² is uniformly heated. if its temperature is increased by 10⁰C, calculate the increase in volume of the cube.

(Given a = 5.0 * 10^-4 ⁰C^-1).

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alok kumar singh

Contributor-Level 10

Given

6a² = 24

a² = 4

a = 2m

$Δ v = v (γ) Δ T$

$= 3 α v Δ T$

$= 3 * 5 * 1 0^{- 4} * (a^{3}) * 10$

$= 3 * 5 * 1 0^{- 4} * 8 * 10$

$\begin{array}{l} = 120 * 1 0^{3} \\ = 1.2 * 1 0^{5} c m^{3} \end{array}$

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New answer posted

a year ago

Physics Laws of Motion

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 $μ$ . The beaker will revolve with the disc if :

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alok kumar singh

Contributor-Level 10

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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a year ago

Physics Motion in a Straight Line

Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by X_p (t) = at + bt² and X_Q (t) = ft – t². At what time, both the buses have same velocity?

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alok kumar singh

Contributor-Level 10

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ -(i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ -(ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$ &

...more

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ -(i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ -(ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

less

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New answer posted

a year ago

Physics 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

a year ago

Physics Class 11th

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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New answer posted

a year ago

Physics Work, Energy and Power

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

0 Follower 4 Views

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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New answer posted

a year ago

Physics 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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New answer posted

a year ago

Physics 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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New answer posted

a year ago

Physics Class 12th

When AC voltage is applied to an inductor, current lags behind voltage by angle of?

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Vikash Kumar Vishwakarma

Contributor-Level 10

The current lags the voltage by an angle of 90 Degree (? /2 radians). This is due to the inductor opposing the change in current.

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