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Locus of the midpoint of any focal chord of \mathrm{y}^2=4 \mathrm{ax} is
 

Option: 1

\mathrm{y}^2=\mathrm{a}(\mathrm{x}-2 \mathrm{a})
 


Option: 2

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-2 \mathrm{a})

 


Option: 3

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-\mathrm{a})
 


Option: 4

\text{None of these}


Let the midpoint be \mathrm{P}(\mathrm{h}, \mathrm{k}). Equation of this chord is \mathrm{\mathrm{T}=\mathrm{S}_1}. i.e., \mathrm{\mathrm{yk}-2 \mathrm{a}(\mathrm{x}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. It must pass through (\mathrm{a}, 0)

\mathrm{\Rightarrow 2 \mathrm{a}(\mathrm{a}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. Thus required locus is \mathrm{\mathrm{y}^2=2 \mathrm{ax}-2 \mathrm{a}^2.}

Hence option 2 is correct.

 

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Posted by

SANGALDEEP SINGH

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

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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

balda gayathri

No. of transition state in given figure

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


2

 

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Posted by

Mura

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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 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

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 N2  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

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Which of the following is an anionic detergent?
Option: 1 Sodium stearate
Option: 2 Sodium lauryl sulphate
Option: 3 Cetyltrimethyl ammonium bromide
Option: 4 Glyceryl oleate

As we have learned

Phenol formation from Benzenesulphonic acid -

Acidification of sodium salt gives phenol.

- wherein

C_{6}H_{5}SO_{3}Na+NaOH\xrightarrow[HCl]{fuse}C_{6}H_{5}-OH+Na_{2}SO_{3}

 

 

 Sodium lauryl sulphate has anionic charge 

 

 

 

 

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Posted by

Ritika Jonwal

The absolute configuration of is :
Option: 1 (2R, 3S)
Option: 2 (2S, 3R)
Option: 3 (2S, 3S)
Option: 4 (2R, 3R)
 

As we learnt ,

 

Chiral Carbon -

Those carbon on which four different groups are present.

- wherein

 

 

For C2 \rightarrow 

\Rightarrow This rotation suggests R but the least prior group is at horizontal position so the configuration is R.

For C3 \rightarrow 

\Rightarrow This rotation suggests S but the least prior group is at horizontal position so the configuration is S.

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Posted by

Ritika Jonwal

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The equilibrium constant at 298 K for a reaction A+BC+D is 100.  If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L−1) will be :
Option: 1  0.182  
Option: 2  0.818  
Option: 3  0.818  
Option: 4  1.182  
 

As we dicussed in the concept

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

 For reaction              A+B\rightleftharpoons C+D

Initial concentration IM      IM       IM       IM

At equilibarium if degree of dissociation is A+B\rightleftharpoons C+D\alpha then

A+B\rightleftharpoons C+D

1-\alpha   1-\alpha      H\alpha   1+\alpha

K_{c}=\frac{(1+\alpha)^{2}}{(1-\alpha)^{2}}=100

10=\frac{1+\alpha}{1-\alpha }

10-10\alpha =1+\alpha

9=11\alpha

\alpha =\frac{9}{11}

Concentration of D is 1+\alpha=I+\frac{9}{11}

=1.818

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Posted by

Ritika Jonwal

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively.  The heat of formation (in kJ) of carbon monoxide per mole is:
Option: 1 110.5
Option: 2 676.5
Option: 3 -676.5
Option: 4 -110
 

\mathrm{C_s+O_2_g \rightarrow CO_2_g \: \: \: \: \Delta H= -393.5 KJmol^{-1}}

CO_{\left ( g \right )}+\frac{1}{2}O_{2} \: _{\left ( g \right )}\rightarrow CO_{2} \:_{(g)} \: \: \: \: \: \Delta H=-283.5 kJmol^{-1}

C_{\left ( s \right )}+\frac{1}{2}O_{2} \:_{\left ( g \right )}\rightarrow CO_{\left ( g \right )}

\therefore \Delta H= -393.5+283.5

            = -110.0\ kJmol^{-1}

Therefore, Option(4) is correct

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Posted by

Ritika Jonwal

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