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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

The following is the biochemical pathway for purple pigment production in flowers of sweet pea:

Colorless precursor 1 —- Allele A—->  Colorless precursor 2 —--- Allele B—---> Purple pigment

Recessive mutation of either gene A or B leads to the formation of white flowers. A cross is made between two parents with the genotype: AaBb × aabb. Considering that the two genes are not linked, the phenotypes of the expected progenies are:

 

Option: 1

9 purple : 7 white


Option: 2

3 white : 1 purple


Option: 3

1 purple : 1 white


Option: 4

9 purple : 6 light purple : 1 white


3(white):1(purple)

In case of 'duplicate recessive epistasis homozygous recessive gene masks the other gene i.e., aa is epistatic to A or a || bb is epistatic to B or b

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Posted by

Riya

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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

A random variable X has the following probability distribution:          X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5 P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2} Then P(X>2) is equal to: 
Option: 1 \frac{7}{12}
Option: 2 \frac{23}{36}
Option: 3 \frac{1}{36}
Option: 4 \frac{1}{6}
 

Option 2

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Posted by

kpvalli2725@gmail.com

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Let a_{n} be the nth term of a G.P. of positive terms. If \sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n is equal to :   
Option: 1 300
Option: 2 175
Option: 3 225
Option: 4 150
 

150

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Posted by

Nalla mahalakshmi

If x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta , for 0<\theta < \frac{\pi }{4}, then
Option: 1 y(1+x)=1
Option: 2 x(1-y)=1
Option: 3 y(1-x)=1
Option: 4 x(1+y)=1
 

x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........

y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......

Use \text S_{\infty}=\frac{1}{1-r}

{x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}

\Rightarrow (1-x)= \sin ^{2} \theta

\Rightarrow y(1-x)=1

Correct Option (3)

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Posted by

avinash.dongre

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If z be a complex number satisfying \left | Re\left ( z \right ) \right |+\left | Im(z) \right |=4, then \left | z \right | cannot be : 
Option: 1 \sqrt{7}
 
Option: 2 \sqrt{\frac{17}{2}}
 
Option: 3 \sqrt{10}
 
Option: 4 \sqrt{8}
 
 

Option d

 

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Posted by

Shravani.D.K

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

Let \left [ t \right ] denote the greatest integer \leq t and \lim_{x\rightarrow 0}x\left [ \frac{4}{x} \right ]=A. Then the function, f\left ( x \right )=\left [ x^{2} \right ]\sin \left ( \pi x \right ) is discontinuous, when x is equal to: 
Option: 1 \sqrt{A+1}
 
Option: 2 \sqrt{A}
 
Option: 3 \sqrt{A+5}
 
Option: 4 \sqrt{A+21}
 
 

 

 

 

\\\lim _{x \rightarrow 0} x\left(\frac{4}{x}-\left\{\frac{4}{x}\right\}\right)=A \quad \Rightarrow \lim _{x \rightarrow 0} 4-x\left\{\frac{4}{x}\right\}=A\\\Rightarrow 4-0=A

\\\text { (1) } x=\sqrt{A+1} \Rightarrow x=\sqrt{5} \Rightarrow \text { discontinuous }\\\text { (2) } x=\sqrt{A+21} \Rightarrow x=5 \Rightarrow \text { continuous }\\\text { (3) } x=\sqrt{A} \Rightarrow x=2 \Rightarrow \text { continuous }\\\text { (4) } x=\sqrt{A+5} \Rightarrow x=3 \Rightarrow \text { continuous }

Correct Option (1)

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avinash.dongre

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