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Class 11th Physics

Match List-I with List-II:

List – I List – II

(a) Magnetic Induction

...more

Match List-I with List-II:

List – I List – II

(a) Magnetic Induction (i) ML²T^-²A^-¹

(b) Magnetic Flux (ii) $M^{0} L^{- 1} A$

(c) Magnetic Permeability (iii) $M T^{- 2} A^{- 1}$

(d) Magnetization (iv) MLT^-²A^-²

Choose the most appropriate answer form the options given below:

less

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alok kumar singh

Contributor-Level 10

$[B] = M T^{- 2} A^{- 1}$

$[?] = M L^{2} T^{- 2} A^{- 1}$

$[μ] = M L T^{- 2} A^{- 2}$

$[M] = M^{0} L^{- 1} A$

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5 months ago

Class 11th Thermodynamics

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5 * 10³ J?

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Vishal Baghel

Contributor-Level 10

$\frac{Δ U}{t} = \frac{6000}{60} - 90$

= 10 J

$t = \frac{Δ U}{10} = \frac{2.5 * 1 0^{3}}{10} = 2.5 * 1 0^{2}$ second

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Class 11th Physics

Inside a uniform spherical shell:

(a) the gravitation field is zero (b) the gravitational potential is zero

(c) the gravitation field is same everywhere (d) the gravitation potential is same everywhere

(e) all the above

Choose the most appropriate answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Inside a uniform spherical shell, electric field is zero every where & electric potential is constant but not zero, every where

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5 months ago

Class 11th Physics

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 :

0 Follower 7 Views

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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5 months ago

Class 11th Physics Motion in Plane

The angle between vector $[\vec{A}] a n d (\vec{A} - \vec{B}) i s :$

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alok kumar singh

Contributor-Level 10

$t a n β = \frac{B s i n 60 °}{A - B c o s 60 °}$

$\Rightarrow β = t a n^{- 1} (\frac{\sqrt[]{3} B}{2 A - B})$

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5 months ago

Class 11th Units and Measurement

If E, L, M and G denote the quantities an energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL²M^-⁵G^-² are:-

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Vishal Baghel

Contributor-Level 10

$P = E L^{2} M^{- 5} G^{- 2}$

$= [M L^{2} T^{- 2}] {[M L^{2} T^{- 1}]}^{2} [M^{- 5}] {[M^{- 1} L^{3} T^{- 2}]}^{- 2}$

$= M^{0} L^{0} T^{0}$

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5 months ago

Class 11th Physics

A refrigerator consumes an average 35W power to operate between temperature -10°C to 25°C. If there is no loss of energy then how much average heat per second does it transfer?

0 Follower 138 Views

alok kumar singh

Contributor-Level 10

$\frac{Q_{H}}{Q_{L}} = \frac{T_{H}}{T_{L}}$

$\Rightarrow Q_{L} = \frac{T_{L}}{T_{H} - T_{L}} * (Q_{H} - Q_{L})$

$\Rightarrow \frac{d Q_{L}}{d t} = \frac{T_{L}}{T_{H} - T_{L}} . \frac{d (Q_{H} - Q_{L})}{d t}$

$= \frac{273 - 10}{25 - (- 10)} * 35$

$= 263 J s^{- 1}$

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5 months ago

Class 11th Motion in a Plane

The magnitude of vectors $\vec{O} A, \vec{O} B a n d \vec{O} C$ in the given figure are equal. The direction of $\vec{O} A + \vec{O} B - \vec{O} C$ with x-axis will be:-

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Vishal Baghel

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

Class 11th Units and Measurement

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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Vishal Baghel

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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New answer posted

5 months ago

Class 11th Physics

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 :

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

F. B. D of Beaker

By NLM2

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

0 0

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