datavorous/entrance-exam-dataset huggingface dataset:
In the binary equation $(1 \mathrm{pl} 01)_{2}+(10 \mathrm{ql})_{2}=(100 \mathrm{r} 00)_{2}$
where $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ are binary digits, what are the possible values of $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ respectively?

Options:
A $0,1,0$
B $1,1,0$
C 0,0,1
D $1,0,1$

The correct answer is:
$0,1,0$

Solution:
$\quad(1 \mathrm{p} 101)_{2}+(10 \mathrm{q} 1)_{2}=(100 \mathrm{r} 00)_{2}$
$\Rightarrow\left(1 \times 2^{4}+\mathrm{p} \times 2^{3}+1 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0}\right)$
$+\left(1 \times 2^{3}+0 \times 2^{2}+q \times 2^{1}+1 \times 2^{0}\right)$
$=1 \times 2^{5}+0+0+\mathrm{r} \times 2^{2}+0+0$
$\Rightarrow 16+8 \mathrm{p}+4+1+8+2 \mathrm{q}+1=32+4 \mathrm{r}$
$\Rightarrow 30+8 p+2 q=32+4 r$
$\Rightarrow 8 \mathrm{p}+2 \mathrm{q}=2+4 \mathrm{r}$
from options, substitute $\mathrm{p}=0, \mathrm{q}=1, \mathrm{r}=0$ we get
$0+2(1)=2+0 \Rightarrow 2=2$.

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 1)
In the biological treatment for sewage, microbes are used for the reduction of the organic matter in the sewage water which are held together by fungal filament to form mesh like structures. These are called as:

Options:
A Activated sludge
B Aerobic process
C Flocs
D Anaerobic sludge

The correct answer is:
Flocs

Solution:
In the wastewater treatment process, the primary effluent is passed into aeration tanks for the secondary treatment. This is also called as biological treatment. The aeration tanks are constantly agitated and the air is pumped; this allows for vigorous growth of aerobic microbes which help in the treatment. Such microbial growth is called flocs which form a mesh due to the association of fungal filaments with bacteria.

Topic: Biology, Microbes in Human Welfare, NEET
In the carbylamine test, why do foul-smelling compounds result from the reaction of primary amines with chloroform and ethanolic potassium hydroxide?

Options:
A Chloroform emits a foul odour when heated.
B Potassium hydroxide produces a strong odour.
C The isocyanide or carbylamine products are foul-smelling.
D Secondary amines produce foul-smelling compounds.

The correct answer is:
The isocyanide or carbylamine products are foul-smelling.

Solution:
The foul odour in the carbylamine test is due to the formation of isocyanides or carbylamines, which are inherently foul-smelling compounds.

Topic: Chemistry, Amines, NEET
In the complex, potassium pentacyanonitrosylvandate (0), the number of potassium atoms per molecule and the coordination number of central metal ion are respectively

Options:
A 5, 5
B 4, 6
C 5, 6
D 4, 4

The correct answer is:
5, 6

Solution:
Here in the complex, potassium pentacyanonitrosylvandate (0), the number of potassium atoms per molecule and the coordination number of central metal ion are 5, 6 respectively as oxidation number of V in this complex is '0' thus the charge on pentacyanonitrosylvandate is -5. Thus formula of the complex is K5[V(CN)5NO] .

Topic: Chemistry, Coordination Compounds, NEET
In the context of a chemical reaction in a beaker, what is considered the "system"?

Options:
A The room where the beaker is kept
B The reaction mixture inside the beaker
C The boundary that separates the beaker from the surroundings
D The beaker itself

The correct answer is:
The reaction mixture inside the beaker

Solution:
In this scenario, the system is defined as the beaker containing the reaction mixture.

Topic: Chemistry, Thermodynamics (C), NEET
In the electrolysis of aqueous sodium chloride solution, which of the half cell reaction will occur
at anode?

Options:
A \( N a^{+}(a q)+e^{-} \rightarrow N a(s) ; E_{\text {cell }}^{\circ}=-2.71 \) volts
B \( 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} ; \mathrm{E}_{\text {cell }}^{\circ}=1.23 \) \$volts
C \( H^{+}(a q)+e^{-} \rightarrow \frac{1}{2} H_{2} ; E_{\text {cell }}^{\circ}=0.00 \) volts
D \( \mathrm{Cl}^{-}(a q) \rightarrow \frac{1}{2} \mathrm{Cl}_{2}+e^{-} ; E_{\text {cell }}^{\circ}=1.36 \) volts

The correct answer is:
\( \mathrm{Cl}^{-}(a q) \rightarrow \frac{1}{2} \mathrm{Cl}_{2}+e^{-} ; E_{\text {cell }}^{\circ}=1.36 \) volts

Solution:
In water sodium chloride ionizes as
\( N a C l(a q) \rightleftharpoons N a^{+}(a q)+C l^{-}(a q) \)
water also dissociates into ions, though to very slight degree as,
\( \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \)
Thus, the aqueous solution of sodium chloride contains \( \mathrm{N} a^{+}, \mathrm{H}^{+}, \mathrm{OH}^{-} \)and \( \mathrm{Cl}^{-} \)ions. When electric current is passed
through the solution, \( \mathrm{Cl}_{2} \) gas evolved at anode and hydrogen is evolved at the cathode. The resulting solution contains
\( \mathrm{Na}^{+} \)and \( \mathrm{OH}^{-} \)ions. Oxidation of \( \mathrm{Cl}^{-} \)ions and \( \mathrm{H}_{2} \mathrm{O} \) is possible at anode but, standard electrode potential of \( \mathrm{Cl}^{-} \)is
more, therefore, \( \mathrm{Cl}^{-} \)will oxidized at anode.

Topic: Chemistry, p Block Elements (Group 13 & 14), KCET, KCET 2017
In the expansion of $(1+a x)^{n}$, the urst three terms are respectively 1, $12 \mathrm{x}$ and $64 \mathrm{x}^{2}$. What is n equal to?

Options:
A 6
B 9
C 10
D 12

The correct answer is:
9

Solution:
The first three terms in expansion of $(1+\mathrm{ax})^{\mathrm{n}}$ are ${ }^{\mathrm{n}} \mathrm{C}_{0},{ }^{\mathrm{n}} \mathrm{C}_{1} \mathrm{ax},{ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{a}^{2} \mathrm{x}^{2}$
Given, ${ }^{1} \mathrm{C}_{0}=1 ;{ }^{\mathrm{n}} \mathrm{C}_{1} \mathrm{ax}=12 \mathrm{x} ;{ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{a}^{2} \mathrm{x}^{2}=64 \mathrm{x}^{2}$
$\Rightarrow \operatorname{nax}=12 \mathrm{x} ; \frac{n(n-1)}{2} a^{2}=64$
$\Rightarrow \mathrm{na}=12 \Rightarrow a=\frac{12}{n}$
$\therefore \frac{n(n-1)}{2} a^{2}=64 \Rightarrow \frac{n(n-1)}{2} \times \frac{144}{n^{2}}=64$
$\Rightarrow \frac{n-1}{n}=\frac{64 \times 2}{144}=\frac{8}{9}$
$\therefore \mathrm{n}=9$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2019 (Phase 1)
In the expansion of $(1+\mathrm{x})^{43}$, if the coefficients of $(2 \mathrm{r}+1)^{\text {th }}$ and $(\mathrm{r}+2)^{\text {th }}$ terms are equal, then what is the value of $\mathrm{r}(\mathrm{r} \neq 1)$ ?

Options:
A 5
B 14
C 21
D 22

The correct answer is:
14

Solution:
Given, in the expansion $\mathrm{g}(1+\mathrm{x})^{43}$, coefficients of $(2 \mathrm{r}+1)^{\text {th }}$ term and $(\mathrm{r}+2)^{\mathrm{th}}$ term are equal.
Coefficient of $(2 \mathrm{r}+1)^{\text {th }}$ term $=\mathrm{n}_{\mathrm{C}_{2 \mathrm{r}}}$
Coefficient of $(\mathrm{r}+2)^{\text {th }}$ term $=\mathrm{n}_{\mathrm{C}_{\mathrm{r}+1}}$
$\mathrm{n}_{\mathrm{C}_{2 \mathrm{r}}}=\mathrm{n}_{\mathrm{C}_{\mathrm{r}+1}}$
$\Rightarrow 43_{\mathrm{C}_{2 \mathrm{r}}}=43_{\mathrm{C}_{\mathrm{r}+1}}$
$(\because \mathrm{n}=43)$
$\Rightarrow 2 \mathrm{r}+\mathrm{r}+1=43$
$\Rightarrow 3 r+1=43$
$\Rightarrow 3 \mathrm{r}+42 \Rightarrow \mathrm{r}=14$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2018 (Phase 1)
In the expansion of $(1+x)^{50}$, the sum of the coefficients of $\begin{array}{ll}\text { odd powers of } \mathrm{x} \text { is } & \end{array}$

Options:
A $2^{26}$
B $2^{49}$
C $2^{50}$
D $2^{51}$

The correct answer is:
$2^{49}$

Solution:
Sum of odd terms of expansion $(\mathrm{a}+\mathrm{b})^{\mathrm{n}}$ is $\frac{1}{2} \cdot 2^{\mathrm{n}}$.
$\therefore$ Sum of odd terms of expansion $(1+\mathrm{x})^{50}$ is $\frac{1}{2} \cdot 2^{50}$
$=2^{-1} .2^{50}=2^{49}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2017 (Phase 2)
In the expansion of $[1+\mathrm{x}]^{\mathrm{n}}$, what is the sum of even binomial coefficients?

Options:
A $2^{\mathrm{n}}$
B $2^{n-1}$
C $2^{n+1}$
D None of the above

The correct answer is:
$2^{n-1}$

Solution:
Sum of all binomial coefficients $=(1+1)^{\mathrm{n}}=2^{\mathrm{n}}$
$\therefore$ Sum of even binomial coefficient $=\frac{2^{\mathrm{n}}}{2}=2^{\mathrm{n}-1}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2012 (Phase 1)
In the expansion of $\left(\sqrt{\mathrm{x}}+\frac{1}{3 \mathrm{x}^{2}}\right)^{10}$ the value of constant term (independent of $\mathrm{x}$ ) is $[2015-I I]$

Options:
A 5
B 8
C 45
D 90

The correct answer is:
5

Solution:
Let $\mathrm{r}^{\text {th }}$ term is independent of $\mathrm{x}$. $\mathrm{T}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \mathrm{y}^{\mathrm{n}-\mathrm{r}}$
$={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{\mathrm{r}}\left(\frac{1}{3 \mathrm{x}^{2}}\right)^{10-\mathrm{r}}$
$={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{3}\right)^{10-\mathrm{r}} \cdot(\sqrt{\mathrm{x}})^{\mathrm{r}}\left(\frac{1}{\mathrm{x}^{2}}\right)^{10-\mathrm{r}}$
Equating the coefficient of $\mathrm{x}$ to zero. $\Rightarrow \mathrm{x}^{\mathrm{r} / 2} \cdot \mathrm{x}^{-2(10-\mathrm{r})}=\mathrm{x}^{0}$
$\Rightarrow \quad \frac{\mathrm{r}}{2}-20+2 \mathrm{r}=0$
$\Rightarrow \quad \frac{5}{2} r=20 \Rightarrow r=8$
Coefficient $={ }^{10} C_{r}\left(\frac{1}{3}\right)^{10-r}$
$={ }^{10} \mathrm{C}_{8}\left(\frac{1}{3}\right)^{10-8}=\frac{10 \times 9}{2} \times \frac{1}{9}=5$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 2)
In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$ where n is a positive integer, the sum of the coefficients of $\mathrm{x}^{5}$ and $\mathrm{x}^{10}$ is $0 .$
What is $n$ equal to ?

Options:
A 5
B 10
C 15
D None of these

The correct answer is:
15

Solution:
$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)
From equation (ii)
${ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}={ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)}$
$\Rightarrow \mathrm{n}=\left(\frac{3 \mathrm{n}}{5}-2\right)+\left(\frac{3 \mathrm{n}}{5}-1\right)$
$\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{y}} \Rightarrow \mathrm{n}=\mathrm{x}+\mathrm{y}\right]$
$\Rightarrow n=\frac{6 n}{5}-3 \Rightarrow \frac{6 n}{5}-n=3$
$\Rightarrow \frac{\mathrm{n}}{5}=3 \quad \therefore \mathrm{n}=15$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 1)
In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$ where n is a positive integer, the sum of the coefficients of $\mathrm{x}^{5}$ and $\mathrm{x}^{10}$ is $0 .$
What is the sum of the coefficients of the two middle terms?

Options:
A 0
B 1
C $-1$
D None of these

The correct answer is:
0

Solution:
$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)
$\mathrm{n}=15$
Total term in the expansion of $\left(\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{2}}\right)^{15}$ is 16 . $\therefore$ middle term $=8^{\text {th }}$ term and $9^{\text {th }}$ term $\mathrm{T}_{8}=\mathrm{T}_{(7+1)}={ }^{15} \mathrm{C}_{7} \cdot(-1)^{7} \cdot \mathrm{x}^{(3 \times 15-5 \times 7)}$
$=-{ }^{15} \mathrm{C}_{7} \cdot \mathrm{x}^{10} \quad$ (from eq. (i))
$\mathrm{T}_{9}=\mathrm{T}_{(8+1)}={ }^{15} \mathrm{C}_{8} \cdot(-1)^{8} \cdot \mathrm{x}^{(3 \times 15-5 \times 8)}$
$=-{ }^{15} \mathrm{C}_{8} \cdot \mathrm{x}^{5} \quad$ (from eq. (ii))
The sum of the coefficients of the two middle terms
$=-{ }^{15} \mathrm{C}_{7}+{ }^{15} \mathrm{C}_{8}=-{ }^{15} \mathrm{C}_{7}+{ }^{15} \mathrm{C}_{7} \cdot\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right]$
$=0$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 1)
In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$ where n is a positive integer, the sum of the coefficients of $\mathrm{x}^{5}$ and $\mathrm{x}^{10}$ is $0 .$
What is the value of the independent term?

Options:
A 5005
B 7200
C $-5005$
D $-7200$

The correct answer is:
$-5005$

Solution:
$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)
For the independent term, put $3 \mathrm{n}-5 \mathrm{r}=0 \quad[$
from eq. $(\mathrm{i})]$
$\Rightarrow 5 \mathrm{r}=3 \mathrm{n}=3 \times 15$
$5 \mathrm{r}=3 \times 3 \times 5$
$\mathrm{r}=9$
Putting the value of r in eq. (i), we get $\mathrm{T}_{9+1}={ }^{15} \mathrm{C}_{9} \cdot(-1)^{9} \cdot \mathrm{x}^{(3 \times 15-5 \times 9)}$
$\Rightarrow \mathrm{T}_{10}=-{ }^{15} \mathrm{C}_{9} \cdot \mathrm{x}^{0}=-{ }^{15} \mathrm{C}_{9}$
$\Rightarrow \mathrm{T}_{10}=-{ }^{15} \mathrm{C}_{6} \quad\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right]$
$=\frac{-15 !}{6 ! 9 !}$
$\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\right]$
$=-5005$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 1)
In the experiment for the determination of the speed of sound in air using the resonance column, the resonates in the fundamental mode, with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction.

Options:
A 0.012 m

B 0.025 m

C 0.05 m

D 0.024 m

The correct answer is:
0.025 m

Solution:
Let âl be the end correction. Given that,
Fundamental tone for a length 0.1m=first overtone for the length 0.35 m
â´v40.1+âl=3v40.35+âl
Solving this equations we get âl=0.025 m=2.5 cm

Topic: Physics, Waves and Sound, NEET
In the following diagram identify the structural and regulatory proteins of muscles?

Options:
A A - troponin ,C - tropomyosin
B A - actin , C - troponin
C B - tropomyosin, A - actin
D A - troponin, C - actin

The correct answer is:
A - troponin, C - actin

Solution:
Actin : Protein that makes up most of the thin myofilaments in a sarcomere muscle fiber.
Troponin : Regulatory protein that binds to actin, tropomyosin, and calcium.
Tropomyosin : Regulatory protein that covers myosin-binding sites to prevent actin from binding to myosin.

Topic: Biology, Locomotion and Movement, NEET
In the following logic circuit, the values of $\mathrm{Y}_1, \mathrm{Y}_2$ and $\mathrm{Y}_3$ are respectively

Options:
A $1,1,1$
B $1,0,0$
C $1,1,0$
D $0,1,1$

The correct answer is:
$1,0,0$

Solution:
No solution. Refer to answer key.

Topic: Physics, Semiconductors, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
In the following pair of halogen compounds, which would undergo SN2 reaction faster?

Options:
A The Primary halide
B The Secondary halide
C Both of them undergo SN2 reaction at the same rate
D Neither of them undergo SN2 reaction

The correct answer is:
The Primary halide

Solution:
Correct Option is : (A)
The Primary halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
In the forest ecosystem, yeasts, molds, and mushrooms are grouped as

Options:
A Producer
B Consumer
C Secondary consumer
D Decomposer

The correct answer is:
Decomposer

Solution:
The decomposers are those living organisms that decompose the dead body of producers, consumers and are known as reducers or transformers or osmotrophs. These are heterotrophic organisms that break up the dead bodies of plants and their waste products. They include fungi like yeasts, molds, and mushrooms and certain bacteria like Bacillus subtilis and Pseudomonas fluorescens etc. The enzymes secreted by these organisms digest the dead organisms and the debris into smaller bits or molecules. These molecules are absorbed by the reducers. After taking energy, the reducers release molecules to the environment as a chemical to be used again by the producers.

Topic: Biology, Ecosystem, NEET
In the fundamental mode, time taken by the wave to reach the closed end of the air filled pipe is $0.01 \mathrm{~s}$. The fundamental frequency is

Options:
A 25
B $12.5$
C 20
D 15

The correct answer is:
25

Solution:
In the fundamental mode,
Frequency, $n=\frac{v}{\lambda}$
$$
\begin{array}{l}
n=\frac{v}{4 l} \\
n=\frac{l}{t \times 4 l} \quad\left[\because v=\frac{l}{t}\right] \\
n=\frac{1}{0.01 \times 4} \\
n=25
\end{array}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2010
In the gaseous phase, what is the order of basicity among amines, including NH3?

Options:
A NH3 > Primary amine > Secondary amine > Tertiary amine
B Tertiary amine > Secondary amine > Primary amine > NH3
C Tertiary amine > Secondary amine > NH3 > Primary amine
D Primary amine > Secondary amine > Tertiary amine > NH3

The correct answer is:
Tertiary amine > Secondary amine > Primary amine > NH3

Solution:
In the gaseous phase, the basicity of amines follows the order mentioned.

Topic: Chemistry, Amines, NEET
In the give diagram, typical chordate plan is given. Identify A and B.

Options:
A Nerve cord and myotomes
B Gut and vertebral column
C Nerve cord and notochord
D Notochord and nerve cord.

The correct answer is:
Nerve cord and notochord

Solution:
Members possess 3 fundamental chordate characters at any stage of life :
1. Notochord - Solid, mesodermal and later on develop into vertebral column.
2. Dorsal hollow nerve cord (or dorsal tubular nerve cord or neural tube) - Hollow, ectodermal and later on develop into CNS (central nervous system).
3. Pharyngeal gill slits.
- Bilaterally symmetrical, organ system level, triploblastic and eucoelomate.

Topic: Biology, Animal Kingdom, NEET
In the given progressive wave $y=5 \sin (100 \pi t-0.4 \pi x)$ where $y$ and $x$ are in metre, $t$ is in second. What is the
(a) amplitude?
(b) wavelength?
(c) frequency?
(d) wave velocity?
(e) particle velocity amplitude?

Options:

None

Solution:
Standard equation of a progressive wave is travelling along positive $x$-direction.
$$
y=a \sin (\omega t-k x+\phi)
$$
As given equation is
$$
y=5 \sin (100 \pi t-0.4 \pi x)
$$
Comparing equation (i) with the standard equation
(a) Amplitude $(a)=5 \mathrm{~m}$
(b) wavelength $(\lambda)$
$$
k=\frac{2 \pi}{\lambda}=0.4 \pi
$$
$\therefore$ Wavelength, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.4 \pi}=\frac{20}{4}=5 \mathrm{~m}$
(c) $\omega=100 \pi$, from eq. (i), $v=\frac{\omega}{2 \pi}$
$$
\therefore \text { Frequency } v=\frac{100 \pi}{2 \pi}=50 \mathrm{~Hz}
$$
(d) Wave velocity
$$
v=v \lambda=50 \times 5=250 \mathrm{~m} / \mathrm{sec}(\because \lambda=5 \mathrm{~m})
$$
(e) Particle (medium) velocity in the direction of amplitude at a distance $x$ from source.
$$
y=5 \sin (100 \pi t-0.4 \pi x)
$$
So, differentiation equation w.r.t. ( $t$ )
Particle velocity
$$
\frac{d y}{d t}=5(100 \pi) \cos [100 \pi t-0.4 \pi x]
$$
For maximum velocity of particle is at its mean position.
So, particle velocity amplitude $\left(\frac{d y}{d t}\right)_{\max }$.
then $\cos [100 \pi t-0.4 \pi x]_{\max }=1$
$\therefore$ Particle velocity amplitude
$$
\begin{aligned}
&=\left(\frac{d y}{d t}\right)_{\max .}=5(100 \pi) \times 1=500 \pi \mathrm{m} / \mathrm{s} \\
&v_{\text {max. }} \text { of medium particle }=500 \pi \mathrm{m} / \mathrm{s} .
\end{aligned}
$$

Topic: Physics, Waves and Sound
In the given reaction

Options:
A Pt/O2
B Pd/O2
C X2/NaOH
D Jones reagent

The correct answer is:
Pd/O2

Solution:
Pt/O2 oxidised -CH2OH into -COOH and it has no effect on secondary hydroxyl group.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the given reaction
R−I + NaCN → DMF R−C≡N
R-I may not be

Options:
A Primary
B Secondary
C Tertiary
D CH3-I

The correct answer is:
Tertiary

Solution:
R−I + NaCN → DMF R−C≡N
Here R-I cannot be tertiary halide as it is SN2 reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
In the given reaction
RCN → (ii)  H 2 O/ H + (i) RMgX (X) , (X) will be

Options:
A Aldehyde
B Ketone
C Tertiary alcohol
D Secondary alcohol

The correct answer is:
Ketone

Solution:
A Cyanide on reaction with Grignard reagent gives Ketone (except HCN)

Topic: Chemistry, Amines, NEET
In the given reaction, the reactivity order of different alcohols is
R-OH+HX→R-X+H2O

Options:
A Primary > Secondary > Tertiary
B Tertiary > Secondary > Primary
C Secondary > Tertiary > Primary
D Secondary > Primary > Tertiary

The correct answer is:
Tertiary > Secondary > Primary

Solution:
R-OX+HX→R-X+H2O , the reaction occurs by SN1 mechanism through carbocation intermediates. Reactivity order of alcohols for this reaction is 3o>2o>1o

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the given reaction

(X) will be

Options:
A 1° -Alcohol
B 2° -Alcohol
C 3° -Alcohol
D Open chain ether

The correct answer is:
2° -Alcohol

Solution:
Here a secondary alcohol (butan-2-ol) is formed as follows

Topic: Chemistry, Alcohols Phenols and Ethers, NEET