Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
Filter By
Clear Filter

All Questions

The product  2^{1/4}\cdot 4^{1/16}\cdot 8^{1/48}\cdot 16^{1/128}\cdot \cdots is equal to :   
Option: 1 2^{1/4}
Option: 2 2
Option: 3 2^{1/2}
Option: 4 1
 

Sum of an infinite GP   

 

If a is the first term and r is the common ratio of a G.P. Then,

\mathrm{S_{\infty}=\frac{a}{1-r}}

S_{\infty} is the sum to infinite terms of the G.P.


Now,

\\2^{1/4}\cdot4^{1/16}\cdot8^{1/48}\ldots=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\ldots}\\\Rightarrow 2^{\frac{\frac{1}{4}}{1-\frac{1}{2}}}=\sqrt 2

View Full Answer(1)
Posted by

avinash.dongre

Which of the following compounds is likely to show both Frenkel abnd Schottky defects in its crystalline form ?
Option: 1 AgBr

Option: 2 CsCl

Option: 3 KBr

Option: 4 ZnS
 

1

View Full Answer(2)
Posted by

Aarchi Jain

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Among the reactions (a)  - (d), the reaction(s) that does / do not occur in the blast furnace during the extraction of iron is / are :

(a)  CaO+SiO_{2}\rightarrow CaSiO_{3} (b)  3Fe_{2}O_{3}+CO\rightarrow 2Fe_{3}O_{4}+CO_{2} (c)  FeO+SiO_{2}\rightarrow FeSiO_{3} (d)  FeO\rightarrow Fe+\frac{1}{2}O_{2}
Option: 1 (C) and (D)

Option:2  (A)

Option: 3 (A) and (D)

Option: 4 (D)
 

The following reactions do not occur in the blast furnace:

C) \mathrm{FeO}+\mathrm{SiO}_{2} \rightarrow \mathrm{FeSiO}_{3}

D) \mathrm{FeO} \rightarrow \mathrm{Fe}+\frac{1}{2} \mathrm{O}_{2}

 

Therefore, Option(1) is correct.

View Full Answer(1)
Posted by

vishal kumar

The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is : (A) Ni(CO)_{4} (B) \left [ Ni(H_{2}O)_{6} \right ]Cl_{2} (C) Na_{2}\left [ Ni(CN)_{4} \right ] (D) PdCl_{2}(PPh_{3})_{2}
Option: 1 (A)\approx (C)< (B)\approx (D)  
  
Option: 2 (C)\approx (D)< (B)< (A)

Option: 3 (C)< (D)< (B)< (A)

Option: 4 (A)\approx (C)\approx (D)< (B)
 

As we have learnt,

[Ni(CO)_4] contains Ni(0) which has a d^{10} configuration having n=0 and \mu=0

[Ni(H_2O)_6]^{2+}contains Ni^{2+} which has a d^8 configuration. In a weak ligand field, it has n=2 and hence has \mu=\sqrt8 \ BM

[Ni(CN)]_4contains Ni^{2+}which has a d^8 configuration. In a strong ligand field, these electrons get paired up and hence n=0 and \mu =0

[Pd(PPh_3)_2Cl_2] contains Pd^{2+}which has a d^8 configuration. These electrons get paired in the presence of strong ligand like PPh_3 and hence, n=0 and correspondingly \mu=0.

Therefore, the correct order of spin only magnetic moment of the complexes is 

(B)>(A)=(C)=(D)

Therefore, Option(4) is correct.

View Full Answer(1)
Posted by

vishal kumar

Crack NEET with "AI Coach"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
neet_ads

A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :
Option: 1 2.40%
Option: 2 5.40%
Option: 3 4.40%
Option: 4 3.40%
 

 

 

As

  • The time period of oscillation of simple pendulum (T)-

 

     T=2\pi \sqrt{\frac{l}{g}}

where 

m=mass of the bob 

l = length of pendulum 

g = acceleration due to gravity.

So

  \begin{array}{l}{\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2}\left(\frac{\Delta \mathrm{g}}{\mathrm{g}}+\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)} \\ {\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{2 \Delta \mathrm{T}}{\mathrm{T}}+\frac{\Delta \mathrm{L}}{\mathrm{L}} ; \quad=2\left(\frac{1}{50}\right)+\frac{0.1}{25.0}} \\ {=4.4 \%}\end{array}

 

Note - We are adding the error just for calculating maximum possible error. 

Hence the correct option is (3).

View Full Answer(1)
Posted by

vishal kumar

If y=y(x) is the solution curve of the differential equation x^{2} \mathrm{~d} y+\left(y-\frac{1}{x}\right) \mathrm{d} x=0 ; x>0$, and $y(1)=1, then y\left(\frac{1}{2}\right) is equal to:
Option: 1 3+e
Option: 2 3+e
Option: 3 3+e
Option: 4 3+e
Option: 5 3-\mathrm{e}
Option: 6 3-\mathrm{e}
Option: 7 3-\mathrm{e}
Option: 8 3-\mathrm{e}
Option: 9 \frac{3}{2}-\frac{1}{\sqrt{e}}
Option: 10 \frac{3}{2}-\frac{1}{\sqrt{e}}
Option: 11 \frac{3}{2}-\frac{1}{\sqrt{e}}
Option: 12 \frac{3}{2}-\frac{1}{\sqrt{e}}
Option: 13 3+\frac{1}{\sqrt{\mathrm{e}}}
Option: 14 3+\frac{1}{\sqrt{\mathrm{e}}}
Option: 15 3+\frac{1}{\sqrt{\mathrm{e}}}
Option: 16 3+\frac{1}{\sqrt{\mathrm{e}}}

x^{2}dy+\left ( y-\frac{1}{x} \right )dx= 0
\Rightarrow x^{2}\frac{dy}{dx}+y= \frac{1}{x}
\Rightarrow\frac{dy}{dx}+\frac{y}{x^{2}}= \frac{1}{x^{3}}
I\cdot F= e^{\int \frac{1}{x^{2}}dx}= e^{-\frac{1}{x}}
\Rightarrow y\cdot e^{-\frac{1}{x}}= \int e^{-\frac{1}{x}}\cdot \frac{1}{x^{3}}dx
Let\, \frac{-1}{x}= t\Rightarrow \frac{1}{x^{2}}dx= dt
\Rightarrow y\cdot e^{-\frac{1}{x}}= -\int e^{t}\cdot t\: dt
\Rightarrow y\cdot e^{-\frac{1}{x}}= -\left [ t\cdot e^{t}-\int e^{t} \; dt\right ]
\Rightarrow ye^{-\frac{1}{x}}= e^{-\frac{1}{x}}+\frac{1}{x}\cdot e^{-\frac{1}{x}}+C
\Rightarrow y= 1+\frac{1}{x}+C\cdot e^{\frac{1}{x}}

As\: y\left ( 1 \right )= 1
\Rightarrow 1= 2+C\cdot e\Rightarrow C= \frac{-1}{e}
\Rightarrow y= 1+\frac{1}{x}-\frac{1}{e}\cdot e^{\frac{1}{x}}
\therefore y\left ( \frac{1}{2} \right )= 1+2-\frac{1}{e}\cdot e^{2}
= 3-e

View Full Answer(1)
Posted by

Kuldeep Maurya

Crack JEE Main with "AI Coach"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
jee_ads

The number of rational terms in the binomial expansion of \left(4^{\frac{1}{4}}+5^{\frac{1}{6}}\right)^{120}is ________
 

21

View Full Answer(2)
Posted by

Anjali

The phase difference between the current and the voltage in series LCR circuit at resonance is
Option: 1 \pi
Option: 2 \frac{\pi}{2}
Option: 3 \frac{\pi}{3}
Option: 4 zero

The phase difference between the current and the voltage in series LCR circuit at resonance is zero

At resonance Z=R

\phi=cos^{-1}\frac{R}{Z}=cos^{-1}\frac{Z}{Z}=0

View Full Answer(1)
Posted by

Safeer PP

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
Option: 1 \mathrm{\pi \ rad}
Option: 2 \mathrm{\frac{3\pi}{2} \ rad}
Option: 3 \mathrm{\frac{\pi}{2} \ rad}
Option: 4 zero

$$\begin{aligned} &\text { SHM is represented by }\ displacement= \mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}\\ &\text { Then, the velocity and acceleration after differentiation will be }\\ &\mathbf{v}=\mathbf{A} \omega \cos \omega \mathbf{t}\\ &\mathrm{a}=-\mathrm{A} \omega^{2} \sin \omega \mathrm{t}=\mathrm{A} \omega^{2} \sin (\omega \mathrm{t}+\pi) \end{aligned}

So the phase difference between x & a= \mathrm{\pi \ rad}

View Full Answer(1)
Posted by

Deependra Verma

A resistance wire connected in the left gap of a metre bridge balances a 10\Omega  resistance in the right gap at a point which divides the bridge wire in the ratio 3:2. If the length of the resistance wire is 1.5 m, then the length of 1\Omega  of the resistance wire is:
Option: 1 1.0\times 10^{-2}m
Option: 2 1.0\times 10^{-1}m
Option: 3 1.5\times 10^{-1}m
Option: 4 1.5\times 10^{-2}m

For a meter bridge

\frac{P}{Q}=\frac{R}{S} =\frac{l}{100-l}

From question

\frac{P}{Q}= \frac{3}{2 } \\ and \ S= 10 \Omega\\ \\ So \ \frac{R}{S}=\frac{3}{2}\\ \\ R=\frac{3}{2}*10=15 \Omega

I.e For 1.5 m long wire its total resistance is 15\Omega

So using the unitary method

The length of 1 \Omega resistance wire will be = L=\frac{1.5}{15}=0.1=1\times10^{-1} \ m

View Full Answer(1)
Posted by

Deependra Verma

Crack NEET with "AI Coach"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
neet_ads
filter_img