Question

${\underline{\bf{Topic : \: \sf{UPSC\:2021}}}}$

1). Photochemical splitting is the result of the reaction between ?

(a) ${\sf{NO_{2}}}$, ${\sf{O_{3}}}$ and peroxyacetyl In the presence of sunlight between nitrates\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ (b) CO, ${\sf{O_{2}}}$ and peroxyacetyl In the presence of sunlight between nitrates

(c) CO, ${\sf{CO_{2}}}$ and ${\sf{NO_{2}}}$ Between, at low temperature

(d) ${\sf{NO_{2}}}$, High concentration of, ${\sf{O_{3}}}$ and CO Between, in the evening​

Answer 1

Answer:

option A

Explanation:

1. Photochemical water splitting comprises three steps . They are absorption of light, generation and separation of charges and redox reactions at the surface as given by the above equations. Reduction of water to hydrogen involves two electrons while the oxidation of water to oxygen involves four electrons.

Answer 2

Answer:

$\huge\red{\mid{\underline{\overline{\texttt{Question}}}\mid}}$

In an Arithmetic progression

$\tt \implies \: t_n=3n+2$

$\tt \implies \: find \: = s_{61}$

$\huge\pink{\mid{\fbox{\tt{Answer↴}}\mid}}$

5795 is the sum of 61 terms of the A.P

$\huge\purple{\mid{\fbox{\tt{Solution}}\mid}}$

GIVEN :-

$\tt \implies \: t_{n} \: = 3n + 2$

TO FIND :-

The sum of first 61 terms $\tt ( \: s_{61})$ ?

FORMULA USED :-

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$\tt \implies \: s_{ {n}^{th} } = \frac{n}{2} [2a + (n - 1)d]$

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where,

➙$\tt ( s_{n ^{th} })$ = sum of $\tt \: {n}^{th}$ of the A.P.

➙n = term number of the number in an A.P

➙a = the First term of the A.P

➙d = common difference between the terms.

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$\tt \implies \: common \: difference(d) \: = t_{2} - t_{1}$

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where,

➙d = common difference between the terms.

➙$\tt \: t_{2} \: =$ second term of the A.P

➙$\tt t_{1} =$ First term of the A.P

CALCULATION :-

As given above ,

$\tt \: t_n=3n+2$

taking n = 1 , then the first term will be :-

$\tt \implies \: t_1=3n+2$

$\tt \implies \: t_1=3(1)+2$

$\tt \implies \: t_1=3+2$

$\tt \implies \: t_1=5$

taking n = 2 ,then the second term will be:-

$\tt \implies \: t_2=3n+2$

$\tt \implies \: t_2=3(2)+2$

$\tt \implies \: t_2=6+2$

$\tt \implies \: t_2=8$

∴The common difference will be :-

$\tt \implies (d) = t_2 - t_1$

substituting the values,

$\tt \implies (d) = 8 - 5$

$\tt \implies (d) = 3$

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Till now we have found :-

➲First term of the A.P = 5

➲Common Difference = 3

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∴sum of the 61 terms of A.P will be :-

$\tt \implies \: s_{ {n}^{th} } = \frac{n}{2} [2a + (n - 1)d]$

$\tt \implies \: s_{ 61} = \frac{n}{2} [2a + (n - 1)d]$

substituting the values,

$\tt \implies \: s_{ 61} = \frac{61}{2} [2(5) + (61 - 1)3]$

$\tt \implies \: s_{ 61} = \frac{61}{2} [10 + (60)3]$

$\tt \implies \: s_{ 61} = \frac{61}{2} (10 + 180)$

$\tt \implies \: s_{ 61} = \frac{61}{2} \times ( 190)$

[Note :- 190 is being divided by 2]

$\tt \implies \: s_{ 61} = 61 \times 95$

$\tt \implies \: s_{ 61} = 5795$

Hence, the sum of 61 terms of the A.P is 5795.