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8 months ago

Class 11th Work, Energy and Power

A car cover AB distance with first one-third at velocity $υ_{1} m s^{- 1},$ second one-third at $υ_{2} m s^{- 1}$ and last one-third at $υ_{3} m s^{- 1} .$ If then the average velocity of the car is__________ ms^-1.

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alok kumar singh

Contributor-Level 10

$t_{1} = \frac{L}{3 V_{1}}$

$t_{2} = \frac{L}{3 V_{2}}$

$t_{3} = \frac{L}{3 V_{3}}$

$V_{a m g} = \frac{L}{t_{1} + t_{2} + t_{3}} = \frac{L}{\frac{L}{3 V_{1}} + \frac{L}{3 V_{2}} + \frac{L}{3 V_{3}}}$

$V_{a m g} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$

$= \frac{3}{\frac{1}{11} + \frac{1}{2 * 11} + \frac{1}{3 * 11}}$

= $\frac{3 * 11 * 6}{6 + 3 + 2}$

= 3 * b

= 18 m/sec

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8 months ago

Class 11th Units and Measurement

The speed of light in media 'A' and 'B' are 2.0 * 10¹⁰ cm/s and 1.5 * 10¹⁰ cm/s respectively. A ray of light enters from the medium B to an incident angle 'q'. If the ray suffers total internal reflection, then

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alok kumar singh

Contributor-Level 10

$\frac{s i n θ_{C}}{V_{B}} = \frac{s i n 90 °}{V_{A}} \Rightarrow s i n θ_{C} = \frac{V_{B}}{V_{A}} = \frac{1.5 * 1 0^{10}}{2.0 * 1 0^{10}} = \frac{3}{4}$

According to question, we can write

$θ > θ = s i n^{- 1} (\frac{3}{4})$

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8 months ago

Class 11th Physics

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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8 months ago

Class 11th Chemistry

50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10^-². (Nearest integer)

(Given : pKa (CH₃-COOH) = 4.76)

log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04

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Payal Gupta

Contributor-Level 10

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$p H = p K a + l o g_{10} \frac{[S]}{[A]}$

$p H = 4.76 + l o g_{10} \frac{2.5}{2.5} = 4.76 = 476 * 1 0^{- 2}$

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8 months ago

Class 11th Physics

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

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alok kumar singh

Contributor-Level 10

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

less

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8 months ago

Class 11th Structure of Atom

A 0.5 percent solution of potassium choride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is__________. (Nearest integer)

(Molal depression constant for water is 1.80 K kg mol^-¹ and molar mass of KCl is 74.6 g mol^-¹)

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Payal Gupta

Contributor-Level 10

0.5 % KCl solution has molality (m) = $\frac{0.5 * 1000}{74.5 * 99.5}$

$K C l_{(a q)} ? K_{(a g)}^{+} + C l_{(a q)}^{-}$

1 - α α α

And I = $(1 - α + α + α) = 1 + α$

$i = \frac{Δ T f}{k f * m} = \frac{0.24 * 74.5 * 99.5}{1.8 * 0.5 * 1000} = 1 + α$

1.976 = 1 + α

$∴ α = 0.976$

% = 97.6%

the nearest 98.

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8 months ago

Class 11th Chemistry Some Basic Concepts of Chemistry

For complete combustion of methanol

$C H_{3} O H (1) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (I)$

The amount of heat produced as measured by bomb calorimeter is 726 kJmol^-¹ at 27°C. The enthalpy of combustion for the reaction is -x kJ mol^-¹, where x is_________

(Nearest integer)

(Given : R = 8.3 J K^-¹ mol^-¹).

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Payal Gupta

Contributor-Level 10

$C H_{3} O H_{(l)} + \frac{3}{2} O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O (l)$

$Δ n = - 0.5$

$Δ H = Δ E + Δ n R T = - 726 - 0.5 * \frac{8.3}{1000} * 300 = - 726 - 1.24 = - 727.24 \approx - 727 k J / m o l$

Hence, x = 727 (the nearest integer)

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8 months ago

Class 11th Electrochemistry

20mL of 0.02 M hypo solution is used for the titration of 10mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of Cu²⁺ is found to be__________* 10^-2 M. [nearest integer]

Given: 2Cu²⁺ + 4I^- ® Cu₂I₂ + I₂

I₂ + $2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}$

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alok kumar singh

Contributor-Level 10

$m e q o f l_{2} = m e q o f N a_{2} S_{2} O_{3} = 20 * 0.002 * 1$

2 * mmol of l₂ = 0.4

mmol of l₂ = 0.2 mmol

mmol of Cu²⁺ = 0.2 * 2 * 10^-3

$[C u^{2 +}] = \frac{0.4 * 1 0^{- 3}}{10 * 1 0^{- 3}} = 0.04 = 4 * 1 0^{- 2} M$

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8 months ago

Class 11th Statistics

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8 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Nine

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