A massless string connects two pulley of masses ' ' and '' 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, ]
Not understanding sir
View Full Answer(2)Calculate the acceleration of block of the following diagram. Assume all surfaces are frictionless . Here m_{1} = 100kg and m_{2} = 50kg
0.33m/s^{2}
0.66m/s^{2}
1m/s^{2}
1.32m/s^{2}
1.32
View Full Answer(4)No. of transition state in given figure
1
2
3
4
2
View Full Answer(2)
When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?
G_{1}/S
G_{2}/M
M
Both G_{2 }M and M
G_{2}/M should be activated as the cell has stalled DNA replication fork.
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In the expansion of , if is the least value of the term independent of when and is the least value of the term independent of when , then the ratio is equal to :
Option: 1
Option: 2
Option: 3
Option: 4
General Term of Binomial Expansion
Term independent of x: It means term containing x^{0},
Now,
Correct option 1
View Full Answer(1)The value of is :
Option: 1
Option: 2
Option: 3
Option: 4
Trigonometric Identities 
Trigonometric Identities
These identities are the equations that hold true regardless of the angle being chosen.

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 (90^{0}  θ) = cos (θ)
cos (90^{0}  θ) = sin (θ)
tan (90^{0}  θ) = cot (θ)
csc (90^{0}  θ) = sec (θ)
sec (90^{0}  θ) = csc (θ)
cot (90^{0}  θ) = tan (θ)
sin (90^{0} + θ) = cos (θ)
cos (90^{0} + θ) =  sin (θ)
tan (90^{0} + θ) =  cot (θ)
csc (90^{0} + θ) = sec (θ)
sec (90^{0} + θ) =  csc (θ)
cot (90^{0} + θ) =  tan (θ)

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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O_{2} 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 C_{4}H_{8}
Option: 2 C_{4}H_{10}
Option: 3 C_{3}H_{6}
Option: 4 C_{3}H_{8}
Volume of N_{2 } in air = 375 × 0.8 = 300 ml
Volume of O_{2} in air = 375 × 0.2 = 75 ml
15ml
0 0 15x 
After combustion total volume
330 = 300 + 15x
x = 2
Volume of O_{2 }used
y = 12
So hydrocarbon is = C_{2}H_{12 }
None of the options matches it therefore it is a BONUS.

Alternatively Solution
15ml
0 0 15x 
Volume of O_{2} used
If further information (i.e., 330 ml) is neglected, option (C_{3}H_{8} ) only satisfy the above equation.
View Full Answer(1) 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 PV^{n}=constant, then n is given by (Here C_{P} and C_{V} are molar specific heat at constant pressure and constant volume, respectively)
Option: 1
Option: 2
Option: 3
Option: 4
For a polytropic preocess
View Full Answer(1)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
where h=2(given)
View Full Answer(1)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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