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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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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

DHFM DVP

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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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

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

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If \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1; then a value of x satisfying y(x)=e is :   
Option: 1 \sqrt{3}\: e
 
Option: 2 \frac{1}{2}\sqrt{3}\: e
 
Option: 3 \sqrt{2}\: e
 
Option: 4 \frac{e}{\sqrt{2}}
 
 

√3e

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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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