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MATHEMATICS

Numerical TypeQuestion 1

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________

CHEMISTRY

Question 2

The bond order and magnetic property of acetylide ion are same as that of

Options:

A)

O2+\mathrm{O}_{2}^{+}

B)

O2−\mathrm{O}_{2}^{-}

C)

N2+\mathrm{N}_{2}^{+}

D)

NO+\mathrm{NO}^{+}
CHEMISTRY

Question 3

When a solution of mixture having two inorganic salts was treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed whereas on treatment with neutral FeCl3\mathrm{FeCl}_{3}, it gave deep red colour which disppeared on boiling and a brown red ppt was formed. The mixture contains :

Options:

A)

SO32−_3^{2-} & C2_2O42−_4^{2-}

B)

CH3COO−& NO3−\mathrm{CH}_{3} \mathrm{COO}^{-} \& ~\mathrm{NO}_{3}^{-}

C)

C2O42−& NO3−\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-} \& ~\mathrm{NO}_{3}^{-}

D)

SO32−& CH3COO−\mathrm{SO}_{3}^{2-} \& ~\mathrm{CH}_{3} \mathrm{COO}^{-}
PHYSICS

Question 4

The energy levels of an atom is shown in figure.

JEE Main 2023 (Online) 25th January Evening Shift Physics - Atoms and Nuclei Question 50 English

Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?

Given (h = 6.62 ×\times 10−34^{-34} Js)

Options:

A)

C

B)

B

C)

A

D)

D
MATHEMATICS

Numerical TypeQuestion 5

Let A=(2−1−110−11−10)A=\left(\begin{array}{rrr}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right) and B=A−IB=A-I. If ω=3i−12\omega=\frac{\sqrt{3} i-1}{2}, then the number of elements in the set⁡{n∈{1,2,…,100}:An+(ωB)n=A+B}\operatorname{set}\left\{n \in\{1,2, \ldots, 100\}: A^{n}+(\omega B)^{n}=A+B\right\} is equal to ____________.

PHYSICS

Question 6

The velocity of a small ball of mass 'm' and density d1, when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is d2, then the viscous force acting on the ball, will be :

Options:

A)

mg(1−d1d2)mg\left( {1 - {{{d_1}} \over {{d_2}}}} \right)

B)

mg(1−d2d1)mg\left( {1 - {{{d_2}} \over {{d_1}}}} \right)

C)

mg(d1d2−1)mg\left( {{{{d_1}} \over {{d_2}}} - 1} \right)

D)

mg(d2d1−1)mg\left( {{{{d_2}} \over {{d_1}}} - 1} \right)