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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.

2

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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.

#### 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

Sodium lauryl sulphate has anionic charge

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#### 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

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

For C3

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

#### 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

are equilibrium concentration

For reaction

Initial concentration IM      IM       IM       IM

At equilibarium if degree of dissociation is then

1-   1-      H   1+

9=

Concentration of D is 1+=I+

=1.818

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#### 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

#### The concentration of fluoride, lead, nitrate, and iron in a water sample from an underground lake was found to be 1 ppm, 40 ppm, 100 ppm and 0.2 ppm, respectively.  This water is unsuitable for drinking due to high concentration of :   Option: 1 Fluoride Option: 10  Lead Option: 19 Nitrate Option: 36 Iron

As we learned in concept

Hard Water -

It contains calcium and Magnesium salt in the form of hydrogen carbonate , chloride and sulphate

- wherein

Hard water does not give Lathers with soap.

Maximum permissible concentration of constituents

nitrate : 50 ppm

Iron : 0.2 ppm

fluoride: 1 ppm

Therefore, option (3) is correct.

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#### Which one of the following statements about water is FALSE? Option: 1 Water is oxidized to oxygen during photosynthesis.. Option: 2 Water can act both as an acid and as a base Option: 3 There is extensive intramolecular hydrogen bonding in the condensed phase Option: 4 Ice formed by heavy water sinks in normal water.

There is intermolecular hydrogen bounding in condensed phase of water and not intramolecular hydrogen bonding.

During Photosynthesis, Water is oxidised to Oxygen according to the reaction

$6CO_2 + 6 H_2O \quad \longrightarrow \quad C_6H_{12}O_6 + 6O_2$

Water can react with both acid as well as base

?