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A massless string connects two pulley of masses ' 2 \mathrm{~kg}' and '1 \mathrm{~kg}' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2


Option: 2

\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2


Option: 3

\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2


Option: 4

\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2


Not understanding sir 

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Posted by

Raju vittal nandi

Calculate the acceleration of block m_1 of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg

 

Option: 1

0.33m/s2


Option: 2

0.66m/s2


Option: 3

1m/s2


Option: 4

1.32m/s2


1.32

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Posted by

balda gayathri

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No. of transition state in given figure

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


2

 

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Posted by

Mura

When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1

G1/S

 

Option: 2

G2/M

 

 

 

Option: 3

M

 

 

Option: 4

Both G2 M and M

 

G2/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

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In the expansion of \left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}, if  L_{1}  is the least value of the term independent of x when \frac{\pi }{8}\leq \theta \leq \frac{\pi }{4} and  L_{2}  is the least value of the term independent of x when \frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}, then the ratio L_{2}:L_{1}  is equal to : 
Option: 1 16:1
Option: 2 8:1
Option: 3 1:8
Option: 4 1:16
 

General Term of Binomial Expansion\left(T_{r+1}\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}

\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}

Term independent of x: It means term containing x0,

 

Now,

\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}\\\text{for r = 8 term is free from 'x' }\\\begin{aligned} &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}\\ &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}\\ &\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \end{aligned}

\\\because \text{Min value of L}_1\;\text{at }\theta=\pi/4\\\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=16 \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\\\\\because \text{Min value of L}_2\;\text{at }\theta=\pi/8\\\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{^{16} \mathrm{C}_{8} \cdot 2^{8}}=16

Correct option 1

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Posted by

avinash.dongre

The value of  \cos ^{3}\left ( \frac{\pi }{8} \right )\cdot \cos \left ( \frac{3\pi }{8} \right )+\sin ^{3}\left ( \frac{\pi }{8} \right )\cdot \sin \left ( \frac{3\pi }{8} \right )  is :
Option: 1 \frac{1}{4}
 
Option: 2 \frac{1}{2\sqrt{2}}
 
Option: 3 \frac{1}{2}
 
Option: 4 \frac{1}{\sqrt{2}}
 
 

 

 

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

 

\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}

-

 

 

 

Allied Angles (Part 1) -

Allied Angles (Part 1)

Two angles or numbers are called allied iff their sum or difference is a multiple of π/2   

  • sin (900 - θ) = cos (θ)

  • cos (900 - θ) = sin (θ)

  • tan (900 - θ) = cot (θ)

  • csc (900 - θ) = sec (θ)          

  • sec (900 - θ) = csc (θ)

  • cot (900 - θ) = tan (θ)

 

  • sin (900 + θ) = cos (θ)

  • cos (900 + θ) = - sin (θ)

  • tan (900 + θ) = - cot (θ)

  • csc (900 + θ) = sec (θ)          

  • sec (900 + θ) = - csc (θ)

  • cot (900 + θ) = - tan (θ)

-

 

 

 

\\\cos ^3\:\frac{\pi }{8}\cos \frac{3\:\pi }{8}+\sin ^3\frac{\pi }{8}\:\sin \frac{3\:\pi }{8}\\\sin\left ( \frac{\pi}{2}-\frac{3\pi}{8} \right )=\cos\left ( \frac{3\pi}{8} \right )=\sin\left ( \frac{\pi}{8} \right )\\\\\\\cos ^3\:\frac{\pi }{8}\sin \frac{\:\pi }{8}+\sin ^3\frac{\pi }{8}\:\cos \frac{\:\pi }{8}\\\sin\frac{\pi}{8}\cos\frac{\pi}{8}\left ( \cos ^2\:\frac{\pi }{8}+\sin ^2\:\frac{\pi }{8} \right )\\\frac{1}{2}\left ( 2\:\sin \frac{\pi }{8}\cos \:\frac{\pi \:}{8} \right )=\frac{1}{2}\:\sin \frac{2\pi }{8}=\frac{1}{2\sqrt2}

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Posted by

avinash.dongre

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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion.  After combustion, the gases occupy 330 mL.  Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
Option: 1  C4H8  
Option: 2  C4H10
Option: 3  C3H6
Option: 4  C3H8
 

Volume of N2  in air = 375 × 0.8 = 300 ml

Volume of O2 in air = 375 × 0.2 = 75 ml
 

C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml                15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

 

After combustion total volume

330 =V_{N_{2}} + V_{CO_{2}}

330 = 300 + 15x 

x = 2 

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

y = 12 

So hydrocarbon is = C2H12

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively  Solution


 C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml              15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

If further information (i.e., 330 ml) is neglected, option (C3H8 ) only satisfy the above equation.

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Posted by

Ritika Jonwal

 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant.  If during  this process the relation of pressure P and volume V is given by PVn=constant,  then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively)
Option: 1  n=\frac{C_{p}}{C_{v}}


Option: 2  n=\frac{C-C_{p}}{C-C_{v}}


Option: 3 n=\frac{C_{p}-C}{C-C_{v}}

Option: 4  n=\frac{C-C_{v}}{C-C_{p}}
 

For a polytropic preocess

c=c_{v}+\frac{R}{1-n} \: or \: \frac{R}{1-n} = c-c_{v}

\Rightarrow 1-n=\frac{R}{c-c_{v}} \: or\: n=1-\frac{R}{c-c_{v}}

\Rightarrow n=\frac{c-\left ( c_{v}+R \right )}{c-c_{v}} = \frac{c-c_{p}}{c-c_{v}}

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Posted by

Ritika Jonwal

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A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure.  The coefficient of friction, between the particle and the rough track equals µ.  The particle is released, from rest, from the point P and it comes to rest at a point R.  The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :
Option: 1  0.2 and 6.5 m  
Option: 3  0.2 and 3.5 m  
Option: 4   0.29 and 6.5 m
 

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

 

decrease in P.E.=P.E.= loss of energy due to friction in PQPQ and QR

\\ m g h=(\mu m g \cos \theta) P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu \times Q R =\mu \cos 30^{\circ} \times 4+\mu \times 2 \sqrt{3} =\mu\left(4 \times \frac{\sqrt{3}}{2}+2 \sqrt{3}\right)\\ h=\mu \times 4 \sqrt{3}\\ \mu=\frac{2}{4 \sqrt{3}}=\frac{1}{2 \sqrt{3}}=0.29where h=2(given)

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Posted by

Ritika Jonwal

In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1  III
Option: 2  I
Option: 3  I and II
Option: 4  III and IV  
 

Geometrical isomerism is not possible at Site I as two identical methyl groups are attached to the same carbon bearing the double bond.

Hence, the answer is Option (2)

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Posted by

vishal kumar

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