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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 GM and M

 

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

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Posted by

Ajit Kumar Dubey

The following system of linear equations  7x+6y-2z=0 3x+4+2z=0 x-2y-6z=0, has 
Option: 1 infinitely many solutions, (x,y,z) satisfying y=2z.
Option: 2 infinitely many solutions, (x,y,z) satisfying x=2z.
Option: 3 no solution
Option: 4 only the trivial solution. 
 

 

 

System of Homogeneous linear equations -

\\\mathrm{Let,} \\\mathrm{a_1x+b_1y +c_1z=0\;\;\; ...(i)} \\\mathrm{a_2x+b_2y +c_2z=0\;\;\; ...(ii)} \\\mathrm{a_3x+b_3y +c_3z=0\;\;\; ...(iii)} \\\mathrm{be \; three\; homogeneous\; equation} \\\mathrm{and \; let\; \Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}}

 

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 non-zero. This solution is called a non-trivial solution.

Explanation: using equation (ii) and (iii), we have 

 

\\\mathrm{\frac{x}{b_2c_3-b_3c_2} = \frac{y}{c_2a_3-c_3a_2}=\frac{z}{a_2b_3-a_3b_2}} \\\\\mathrm{or \;\; \frac{x}{\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}}=\frac{y}{\begin{vmatrix} c_2 &a_2 \\ c_3 & a_3 \end{vmatrix}} = \frac{z}{\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} = k (say \neq 0)} \\\mathrm{\therefore x = k\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}, y = k\begin{vmatrix} c_2& a_2\\ c_3 & a_3 \end{vmatrix} \; and \; z = k\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} \\\mathrm{putting\; these\; value\; in \; equation \; (i), we\; have} \\\mathrm{a_1\left \{ k\begin{vmatrix} b_2 & c_2\\ b_3 & c_3 \end{vmatrix} \right \}+b_1\left \{ k\begin{vmatrix} c_2 & a_2\\ c_3 & a_3 \end{vmatrix} \right \}+c_1\left \{ \begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix} \right \}=0}

 

\\\mathrm{\Rightarrow a_1\begin{vmatrix} b_2 &c_2 \\ b_3 & c_2 \end{vmatrix}-b_1\begin{vmatrix} a_2 & c_2\\ a_3 &c_3 \end{vmatrix}+c_1\begin{vmatrix} a_2 &b_2 \\ a_3 &b_3 \end{vmatrix} = 0 \;\;\;[\because \; k \neq 0]} \\\mathrm{or \;\; \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \; or \; \Delta = 0}

This is the condition for a system have Non-trivial solution.

-

 

 

\begin{aligned} &(1)\;\;7 x+6 y-2 z=0\\ &(2)\;\;3 x+4 y+2 z=0\\ &(3)\;\;x-2 y-6 z=0 \end{aligned}

\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right| =7(-20)-6(-20)-2(-10)=-140+120+20=0

so infinite non-trivial solution exist

now equation (1) + 3 equation (3)

10x - 20z = 0

x = 2z

Correct Option 2

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Posted by

avinash.dongre

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Let a-2b+c=1.  If f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix}, then :   
Option: 1 f(-50)=501
Option: 2 f(-50)=-1
Option: 3 f(50)=1
Option: 4 f(50)=-501
 

 

 

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 ith row with jth row, this operation is denoted by 

        \\\mathrm{R_i \leftrightarrow R_j \;\;or\; R_{ij}}

    ii) The multiplication of ith row by a constant k (k≠0) is denoted by

         \\\mathrm{R_i \leftrightarrow kR_i \;\; or \; k\cdot R_i}    

    iii) The addition of ith row to the elements of jth row multiplied by constant k (k≠0) is denoted by

        \\\mathrm{R_i \leftrightarrow R_i + kR_j \;\; or \; k\cdot R_{ij}}

In the same way, three-column operations can also be defined too.

-

 

 

\\\text { Apply } \mathrm{R}_{1}=\mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\\\Rightarrow f(x)=\left|\begin{array}{ccc}{1} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right| \quad \Rightarrow f(x)=1 \\\quad \Rightarrow f(50)=1

Correct Option (3)

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Posted by

avinash.dongre

If for some \alpha \: and\: \beta in R, the intersection of the following three planes  x+4y-2z=1 x+7y-5z=\beta x+5y+\alpha z=5 is aline in R^{3}, then \alpha +\beta is equal to : 
Option: 1 0
Option: 2 10
Option: 3 -10
Option: 4 2
 

 

 

Cramer’s law -

Cramer’s law for the system of equations in two variables :

\\\mathrm{Let \; a_1x +b_1y = c_1\; and \; a_2x + b_2y = c_2, where} \\\mathrm{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}} \\\mathrm{on \; solving \; this \; equation \; by \; cross \; multiplication, we \; get} \\\mathrm{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}} \\\mathrm{or \; \frac{x}{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}=\frac{y}{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}=\frac{1}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}} \\\mathrm{or \; x=\frac{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}, y=\frac{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}}

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.

 

 

\\\mathrm{so\; let \;us\; consider\; the \; system\; of \;equations} \\\mathrm{a_1x+b_1y +c_1z =d_1 \;\;...(i)} \\\mathrm{a_2x+b_2y +c_2z =d_2\;\;\;...(ii)} \\\mathrm{a_3x+b_3y +c_3z =d_3\;\;\;...(iii)} \\\mathrm{then \; \Delta,\; which\; will\; be \;determinant\; of\; coefficient\; of \;variables, will\; be } \\\mathrm{\Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2& b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}, \Delta_1\; numerator\; of\; x \;is:} \\\mathrm{\Delta_1= \begin{vmatrix} d_1 & b_1 &c_1 \\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3 \end{vmatrix}, similarly\; \Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1\\ a_2 & d_2 & c_2\\ a_3 & d_3 & c_3 \end{vmatrix}\; and \;} \\\mathrm{\Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3 \end{vmatrix}, if \Delta \neq 0\; then\; \;x=\frac{\Delta_1}{\Delta}} \\\mathrm{y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}

 

i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are  \\\mathrm{x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}

 

ii) If ? = 0, and any of \Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0

Then the system of equations is inconsistent and hence no solution exists.

 

iii) If all \Delta =\Delta_1=\Delta_2=\Delta_3= 0 then

System of equations is consistent and dependent and it has an infinite number of solution.

-

 

 

\begin{array}{l}{\Delta=0 \Rightarrow\left|\begin{array}{ccc}{1} & {4} & {-2} \\ {1} & {7} & {-5} \\ {1} & {5} & {\alpha}\end{array}\right|=0} \\ {(7 \alpha+25)-(4 \alpha+10)+(-20+14)=0} \\ {3 \alpha+9=0 \Rightarrow \alpha=-3}\end{array}

 

\begin{array}{l}{\text { Also }\quad D_{3}=0 \Rightarrow\left|\begin{array}{lll}{1} & {4} & {1} \\ {1} & {7} & {\beta} \\ {1} & {5} & {5}\end{array}\right|=0} \\ {1(35-5 \beta)-(15)+1(4 \beta-7)=0} \\ {\beta=13}\end{array}

\alpha+ \beta=10

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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 N 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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 Which one of the following is an oxide ?
Option: 1  KO2
Option: 2  BaO2
Option: 3  SiO2
Option: 4 CsO2  
 

SiO2  - oxide

KO???2? , CsO????2  -  superoxides

BaO????2 -  peroxide

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Posted by

vishal kumar

 The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal : Fe2O3(s)+3 CO(g) 2 Fe(l)+3 CO2(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 CO2
Option: 3 Addition of CO2
Option: 4 Addition of Fe2O3  
 

\\Fe_{2}O_{3}+3CO \leftrightharpoons 2Fe+3CO \\

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 Fe2O???3 will not disturb the equilibrium. 

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Posted by

vishal kumar

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