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.
View Full Answer(1)The following system of linear equations has
Option: 1 infinitely many solutions, satisfying
Option: 2 infinitely many solutions, satisfying
Option: 3 no solution
Option: 4 only the trivial solution.
System of Homogeneous linear equations 
If ? ≠ 0, then x= 0, y = 0, z = 0 is the only solution of the above system. This solution is also known as a trivial solution.
If ? = 0, at least one of x, y and z are nonzero. This solution is called a nontrivial solution.
Explanation: using equation (ii) and (iii), we have
This is the condition for a system have Nontrivial solution.

so infinite nontrivial solution exist
now equation (1) + 3 equation (3)
10x  20z = 0
x = 2z
Correct Option 2
View Full Answer(1)Let If then :
Option: 1
Option: 2
Option: 3
Option: 4
Elementary row operations 
Elementary row operations
Row transformation: Following three types of operation (Transformation) on the rows of a given matrix are known as elementary row operation (transformation).
i) Interchange of i^{th} row with j^{th} row, this operation is denoted by
ii) The multiplication of ith row by a constant k (k≠0) is denoted by
iii) The addition of ith row to the elements of jth row multiplied by constant k (k≠0) is denoted by
In the same way, threecolumn operations can also be defined too.

Correct Option (3)
View Full Answer(1)If for some in R, the intersection of the following three planes is aline in , then is equal to :
Option: 1
Option: 2
Option: 3
Option: 4
Cramer’s law 
Cramer’s law for the system of equations in two variables :
We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.
We can follow this analogy for the system of equations of 3 variable where third in the numerator of the value of z will be of constant and denominator will be formed by the value of coefficients of the variables.
i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are
ii) If ? = 0, and any of
Then the system of equations is inconsistent and hence no solution exists.
iii) If all then
System of equations is consistent and dependent and it has an infinite number of solution.

View Full Answer(1)
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)
View Full Answer(1) Which one of the following is an oxide ?
Option: 1 KO_{2}
Option: 2 BaO_{2}
Option: 3 SiO_{2}
Option: 4 CsO_{2}
SiO_{2}  oxide
KO_{???2}? , CsO_{????2 }  superoxides
BaO_{????2}  peroxide
View Full Answer(1) The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal : Fe_{2}O_{3}(s)+3 CO(g) 2 Fe(l)+3 CO_{2}(g) Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?
Option: 1 Removal of CO
Option: 2 Removal of CO_{2}
Option: 3 Addition of CO_{2}
Option: 4 Addition of Fe_{2}O_{3}
According to Le Chatelier's principle change in concentration by changing the amount of reactant or product affect the equilibrium. However, the addition of solid reactant won't affect the concentration.
Therefore, addition of solid Fe_{2}O_{???3} will not disturb the equilibrium.
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