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3 months ago

Class 11th Laws of Motion

A block of mass 200g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F $= \sqrt[]{x}$ N as shown in figure.

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alok kumar singh

Contributor-Level 10

For equilibrium along the incline plane is given by,

F cos 60° = mg sin 60°

$F = \frac{0.2 * 10 * \sqrt[]{3}}{2 * 1} * 2$

$= 2 \sqrt[]{3} N = \sqrt[]{12} N$

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3 months ago

Class 11th Chemical Bonding and Molecular Structure

Given below are two statements one is labeled as Assertion A and the other is labeled as Reason R:

Assertion A: The amphoteric nature of water is explained by using Lewis acid/base concept

Reason R: Water acts as an acid with NH₃ and as a base with H₂S

In the light of the above statements choose the correct answer form the options given below:

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Vishal Baghel

Contributor-Level 10

$\underset{(a c i d)}{H_{2} S} + \underset{B a s e}{H_{2} O} ? H_{3} O^{+} + H S^{-}$

$\underset{(a c i d)}{H_{2} O} + \underset{(B a s e)}{N H_{3}} ? N H_{4} O H$

With H₂S water acts like base and with NH₃ it acts like acid.

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3 months ago

Class 11th Laws of Motion

A curved in a level road has aradius 75m. The maximum speed of a car turning this curved road can be 30 m/s without skidding.; If radius of curved road is changed to 48m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be…….m/s.

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alok kumar singh

Contributor-Level 10

$f = \frac{m v^{2}}{R}$

\Rightarrow \frac{m v^{2}}{R} = μ N

$\frac{m v^{2}}{R} = μ m g$

$\Rightarrow 30 = \sqrt[]{μ * 75 * 10}$

$μ = \frac{30 * 30}{750} = 1.2$

Now, R = 48 m (new Radius of curvature)

$μ = 1.2$

$V_{m a x} = \sqrt[]{μ R g}$

$= \sqrt[]{1.2 * 48 * 10}$

$= 24 m / s e c$

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New answer posted

3 months ago

Class 11th Periodic Table Elements

The correct order of electron gain enthalpies of Cl, F, Te and Po is

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Vishal Baghel

Contributor-Level 10

Element e^- gain enthalpy

F -333

Cl -349

Te -190

Po &nb

...more

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New answer posted

3 months ago

Class 11th Thermodynamics

At $25 ° C$ and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are -3268 kJ mol^-1 and -1300 kJ mol^-1, respectively. The change in enthalpy for the reaction 3C₂H₃(g) ® C₆H₆(I), is

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Vishal Baghel

Contributor-Level 10

$Δ H = 3 {(Δ H_{c o m b})}_{C_{2} H_{2} (g)} - {(Δ H_{c o m b})}_{C_{6} H_{6} (l)}$

$= - 1300 * 3 - (- 3268) k J / m o l e$

$= (- 3900 + 3268) k J / m o l e$

$= - 632 k J / m o l e$

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3 months ago

Class 11th Units and Measurement

For $z = a^{2} x^{3} y^{\frac{1}{2}},$ where 'a' is a constant. If percentage error in measurement of 'x' and 'y' are 4% and 12% respectively, then the percentage error for 'z' will be…………%

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alok kumar singh

Contributor-Level 10

$\frac{Δ z}{z} * 100 = 3 \frac{Δ x}{x} * 100 + \frac{1}{2} \frac{Δ y}{y} * 100$

$= 3 * 4 + \frac{1}{2} * 12$

$= (12 + 6) %$

$= 18 %$

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3 months ago

Class 11th Physics

The root mean square speed of smoke particles of mass 5 * 10^-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 * 10^-23 JK^-1]

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Payal Gupta

Contributor-Level 10

m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 * 1.38 * 1 0^{- 23} * 300}{5 * 1 0^{- 17}}}$

$= \sqrt[]{18 * 13.8 * 1 0^{- 6}}$

$= 15.7 * 1 0^{- 3} m / s$

$= 15 m m / s$

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New answer posted

3 months ago

Class 11th Physics Thermodynamics

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

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Payal Gupta

Contributor-Level 10

W_total = W_D→E + W_E→F = Area of triangle DEF

= $\frac{1}{2} * 3 * 300 = 450 J .$

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3 months ago

Class 11th Maths

Which of the following statements is a tautology?

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Vishal Baghel

Contributor-Level 10

Truth table

$p q \sim p \sim q (\sim p) \lor q ((\sim p) \lor q) \to p p \to ((\sim p) \lor q) (\sim p) \lor q \to q q \to ((\sim p) \lor q)$

T T F F T T T T &

...more

Truth table

$p q \sim p \sim q (\sim p) \lor q ((\sim p) \lor q) \to p p \to ((\sim p) \lor q) (\sim p) \lor q \to q q \to ((\sim p) \lor q)$

T T F F T T T T T

T F F T F T F T T

F T T F T F T T T

F F T T T F T F T

less

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New answer posted

3 months ago

Class 11th Maths

A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15m. If from a point A on the ground the angle of elevation of R is $60 °$ and the part PR of the tower subtends an angle of $15 °$ at A, then the height of the tower is:

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Vishal Baghel

Contributor-Level 10

From $Δ A P Q$

$\frac{x + 15}{y} = t a n 75 °$ …. (i)

From $Δ R Q A$

$\frac{}{}$ $\frac{15}{y} = t a n 60 °$ …… (ii)

From (i) and (ii)

$= 5 (2 \sqrt[]{3} + 3) m$

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