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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

View Full Answer(2)

Posted by

Mura

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

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 G₂M 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 O₂used

$15\left ( x +\frac{y}{4} \right ) = 75$

$\left ( x +\frac{y}{4} \right ) = 5$

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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Get Answers to all your Questions

All Questions

Locus of the midpoint of any focal chord of is Option: 1 Option: 2 Option: 3 Option: 4

Posted by

SANGALDEEP SINGH

Posted by

Raju vittal nandi

Crack CUET with india's "Best Teachers"

Calculate the acceleration of block of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg Option: 1 0.33m/s2Option: 2 0.66m/s2Option: 3 1m/s2Option: 4 1.32m/s2

Posted by

balda gayathri

No. of transition state in given figure Option: 1 1Option: 2 2Option: 3 3Option: 4 4

Posted by

Mura

JEE Main high-scoring chapters and topics

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

Posted by

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
$\text{None of these}$

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

No. of transition state in given figure

Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

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 G₂M and M

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

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

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

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