Physics

Physics Units and Measurement

Contributor-Level 10

Let's say $| \vec{O A} | = | \vec{O B} | = | \vec{O C} | = l$

$\vec{O A} = l c o s 30 ° \hat{i} + l s i n 30 ° \hat{j}$

$\vec{O A} + \vec{O B} - \vec{O C}$

= $[l c o s 30 ° \hat{i} + l s i n 30 ° \hat{j}] + [l c o s 60 ° \hat{i} + l s i n 60 ° (- \hat{j})] - [l c o s 45 ° (- \hat{i}) + l s i n 45 ° (+ \hat{j})]$

$= l (\frac{\sqrt[]{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt[]{2}}) \hat{i} + l [\frac{1}{2} - \frac{\sqrt[]{3}}{2} - \frac{1}{\sqrt[]{2}}] \hat{j}$

$t a n θ = \frac{l [\frac{1}{2} - \frac{\sqrt[]{3}}{2} - \frac{\sqrt[]{2}}{2}]}{l [\frac{\sqrt[]{3}}{2} + \frac{1}{2} + \frac{\sqrt[]{2}}{2}]} = \frac{1 - \sqrt[]{3} - \sqrt[]{2}}{1 + \sqrt[]{3} + \sqrt[]{2}}$

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5 months ago

In a Screw Gauge, fifth division of the circular scale coincides with reference line when the ratched is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5mm on a complete rotation. For a particular observation the reading on the main scale is 5mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.

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Contributor-Level 10

Pitch = 0.5 mm

Least count = $\frac{p i t c h}{d i v i s i o n} = \frac{0.5 m m}{50} = 0.01 m m$ $\frac{}{} \frac{}{}$

True Reading = 5mm + 20 * 0.01 – 5 * 0.01 = 5.15 mm

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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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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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The rms speed of the molecules of Hydrogen, Oxygen and Carbondioxide at the same temperature are $V_{H}, V_{O} a n d V_{C}$ respectively then:

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Contributor-Level 10

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}}$

$V_{r m s} \propto \frac{1}{\sqrt[]{M}}$ at constant T

$V_{H} : V_{O} : V_{C o_{2}} = \frac{1}{\sqrt[]{1}} : \frac{1}{\sqrt[]{16}} : \frac{1}{\sqrt[]{44}}$

$= 1 : \frac{1}{4} : \frac{1}{2 \sqrt[]{11}}$

$V_{H} > V_{O} > V_{C O_{2}}$

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Physics Class 12th

A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in figure. The approximate variation electric field $\vec{E}$ as a function of distance r from centre O is given by:

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Contributor-Level 10

Assuming shell to be conducting

For $\frac{0 \leq r < R}{q_{i n} = 0},$

E = 0

$\frac{R \leq r < a}{q_{i n} = Q}$

$E = \frac{k Q}{r^{2}}$

$\frac{a \leq r < b}{q_{i n} = Q - Q = 0}$

E = 0

$\frac{r \geq b}{q_{i n} = Q}$

$E = \frac{k Q}{r^{2}}$

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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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Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

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

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

(i) = (ii)

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

$t = \frac{f - α}{2 (β + 1)}$

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Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

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

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

(i) = (ii)

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

$t = \frac{f - α}{2 (β + 1)}$

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

Reason R : Product of said heights.

$h_{1} h_{2} = (\frac{u^{2} s i n^{2} θ}{2 g}) . (\frac{u^{2} c o s^{2} θ}{2 g})$

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Physics Physics Motion in Plane

Contributor-Level 9

First angle, 1 = $R = \frac{u^{2} s i n 2 ?}{g}$ $\frac{^{}}{}$

Another angle (2 = 90 - ) for which range will be

Same as that of 1 =

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

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

$& ? ? ? ?_{2} = 90 ? ?, h_{2} = \frac{u^{2} s i n^{2} ?_{2}}{2 g} = \frac{u^{2} c o s^{2} ?}{2 g}$

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

$h_{1} h_{2} = \frac{1}{4^{2}} R^{2}$

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

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

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Physics Class 12th

A teacher in his physics laboratory allotted an experiment ot determine the resistance (G) of a galvanometer. Students took the observations for $\frac{1}{3}$ deflection in the galvanometer. Which of the below is true for measuring value of G?

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