datavorous/entrance-exam-dataset huggingface dataset:
$\cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)=$

Options:
A $\frac{-1}{8}$
B $\frac{1}{8}$
C $-\frac{3 \sqrt{3}}{8}$
D 1

The correct answer is:
$\frac{-1}{8}$

Solution:
We have,
$$
\cos \frac{\pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7}
$$
Let $A=\frac{\pi}{7} \Rightarrow \pi=7 A$
$$
=\cos A \cos 2 A \cos 4 A=\cos A \cos 2 A \cos 2^2 A
$$
Now, we will use the formula
$$
\begin{aligned}
& \cos A \cos 2 A \cos 2^2 A \cos 2^3 A \ldots \ldots \cos ^{n-1} A=\frac{\sin 2^n A}{2^n \sin A} \\
& \therefore \cos A \cos 2 A \cos 2^2 A=\frac{\sin 2^3 A}{2^3 \sin A} \\
& =\frac{\sin 8 A}{8 \sin A}=\frac{\sin (7 A+A)}{8 \sin A}=\frac{\sin (\pi+A)}{8 \sin A} \quad[\because 7 A=\pi] \\
& =\frac{-\sin A}{8 \sin A} \quad[\because \sin (\pi+\theta)=-\sin \theta] \\
& =-\frac{1}{8}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
\( \cos 40^{\circ} \cos 80^{\circ} \cos 160^{\circ}= \)

Options:
A \( \frac{-1}{2} \)
B \( \frac{-1}{4} \)
C \( \frac{-1}{8} \)
D \( \frac{-1}{16} \)

The correct answer is:
\( \frac{-1}{8} \)

Solution:
cos60-20·cos60+20·cos160
=cos20°cos20°×cos°60-20×cos60+20°×cos160°

Now using cosθ·cos60°-θ·cos60°+θ=14cos3θ

=14cos60°×cos180-20°cos20°

⇒14×12×-cos20cos20

= -18

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\cos A=\frac{3}{4} \Rightarrow 32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5 A}{2}\right)=$

Options:
A 7
B 8
C 13
D 11

The correct answer is:
11

Solution:
$\begin{aligned} & 16\left[2 \sin \frac{A}{2} \sin \left(\frac{5 A}{2}\right)\right]=16[\cos 2 A-\cos 3 A] \\ & =16\left[2 \cos ^2 A-1-\left(4 \cos ^3 A-3 \cos A\right)\right] \\ & =16\left[2\left(\frac{3}{4}\right)^2-1-\left(4 \times \frac{27}{64}-3 \times \frac{3}{4}\right)\right] \\ & =16\left[\frac{1}{8}-\left(\frac{27}{16}-\frac{9}{4}\right)\right] \\ & =16\left[\frac{2-27+36}{16}\right] \\ & =11\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

Options:
A $\frac{\sin 2^n A}{2^n \sin A}$
B $\frac{2^n \sin 2^n A}{\sin A}$
C $\frac{2^n \sin A}{\sin 2^n A}$
D $\frac{\sin A}{2^n \sin 2^n A}$

The correct answer is:
$\frac{\sin 2^n A}{2^n \sin A}$

Solution:
It is a standard result.
$\begin{gathered}
\cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A \\
=\frac{\sin 2^n A}{2^n \sin A}
\end{gathered}$

Topic: Mathematics, Trigonometric Equations, JEE Main
$\cos A+\cos \left(240^{\circ}+A\right)+\cos \left(240^{\circ}-A\right)=$

Options:
A $\cos A$
B 0
C $\sqrt{3} \sin A$
D $\sqrt{3} \cos A$

The correct answer is:
0

Solution:
$\begin{aligned}
\cos A+\cos \left(240^{\circ}\right. & +A)+\cos \left(240^{\circ}-A\right) \\
& =\cos A+2 \cos 240^{\circ} \cos A \\
& =\cos A\left\{1+2 \cos \left(180^{\circ}+60^{\circ}\right)\right\}=\cos A\left\{1+2\left(-\frac{1}{2}\right)\right\}
\end{aligned}$
$=0$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2=$

Options:
A $4 \cos ^2 \frac{\alpha-\beta}{2}$
B $4 \sin ^2 \frac{\alpha-\beta}{2}$
C $4 \cos ^2 \frac{\alpha+\beta}{2}$
D $4 \sin ^2 \frac{\alpha+\beta}{2}$

The correct answer is:
$4 \cos ^2 \frac{\alpha-\beta}{2}$

Solution:
$\begin{aligned} & \text { Soln: (a) }(\cos \alpha+\cos \beta)^2+(\sin \alpha+\sin \beta)^2 \\ & =\cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta+\sin ^2 \alpha+ \\ & \sin ^2 \beta+2 \sin \alpha \sin \beta \\ & =2\{1+\cos (\alpha-\beta)\}=4 \cos ^2\left(\frac{\alpha-\beta}{2}\right) \\ & \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\cos \left[\lim _{x \rightarrow \infty} \frac{2 \pi|x|+\pi x}{|x|-3 x}+\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2} \cos ^2 x\right)}{x^2}\right]$

Options:
A 1
B -1
C 0
D $\frac{1}{\sqrt{2}}$

The correct answer is:
-1

Solution:
$\begin{aligned} & \text { } \cos \left[\lim _{x \rightarrow \infty} \frac{2 \pi|x|+\pi x}{|x|-3 x}+\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2} \cos ^2 x\right)}{x^2}\right] \\ & =\cos \left[\lim _{x \rightarrow \infty} \frac{2 \pi x+\pi x}{x-3 x}+\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}\left(1-\sin ^2 x\right)\right)}{x^2}\right] \\ & =\cos \left[\lim _{x \rightarrow \infty} \frac{3 \pi x}{-2 x}+\lim _{x \rightarrow 0}\left(\frac{\sin \left(\frac{\pi}{2} \sin ^2 x\right)}{\left(\frac{\pi}{2} \sin ^2 x\right)} \times \frac{\frac{\pi}{2} \sin ^2 x}{x^2}\right)\right] \\ & =\cos \left[-\frac{3 \pi}{2}+\frac{\pi}{2}\right]=\cos (-\pi)=-1 .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\int \cos \sqrt{x} d x=$

Options:
A $2[\sqrt{x} \sin \sqrt{x}+\cos \sqrt{x}]+c$
B $2[\sqrt{x} \sin \sqrt{x}-\cos \sqrt{x}]+c$
C $2[\cos \sqrt{x}-\sqrt{x} \sin \sqrt{x}]+c$
D $-2[\sqrt{x} \sin \sqrt{x}+\cos \sqrt{x}]+c$

The correct answer is:
$2[\sqrt{x} \sin \sqrt{x}+\cos \sqrt{x}]+c$

Solution:
$\begin{aligned} & \text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow d x=2 t d t \text {, } \\ & \int 2 t \cdot \cos t d t=2\left[t \cdot \sin t-\int \sin t d t\right] \\ & =2 t \sin t+2 \cos t=2[\sqrt{x} \sin \sqrt{x}+\cos \sqrt{x}]+c .\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
cosec2cot-15+cos-145 is equal to:

Options:
A 6556
B 7556
C 6533
D 5633

The correct answer is:
6556

Solution:
Given, cosec2tan-115+tan-134

=cosectan-12×151-125+tan-134

=cosectan-1512+tan-134

=cosectan-1512+341-512×34

=cosectan-15633=6556

option 1

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (25 Feb Shift 2)
$\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]=$

Options:
A $\pi-X$
B $2 \pi-x$
C $\frac{x}{2}$
D $\pi-\frac{x}{2}$

The correct answer is:
$\pi-\frac{x}{2}$

Solution:
$\begin{aligned}
& \cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right] \\
& =\cot ^{-1}\left[\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}-\sqrt{1+\sin x})} \cdot \frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}\right] \\
& =\cot ^{-1}\left[\frac{(1-\sin x)+(1+\sin x)+2 \sqrt{1-\sin ^2 x}}{(1-\sin x)-(1+\sin x)}\right] \\
& =\cot ^{-1}\left[\frac{2(1+\cos x)}{-2 \sin x}\right]=\cot ^{-1}\left[-\frac{2 \cos ^2(x / 2)}{2 \sin (x / 2) \cos (x / 2)}\right] \\
& =\cot ^{-1}\left(-\cot \frac{x}{2}\right)=\cot ^{-1}\left[\cot \left(\pi-\frac{x}{2}\right)\right]=\pi-\frac{x}{2}
\end{aligned}$
Trick : Put $ x=\frac{\pi}{4}$, so that the expression becomes
$\begin{aligned}
& \cot ^{-1}\left[\frac{\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}}\right] \\
& =\cot ^{-1}\left[\frac{\sqrt{2}-1+\sqrt{2}+1+2 \sqrt{2-1}}{\sqrt{2}-1-\sqrt{2}-1}\right]
\end{aligned}$
$=\cot ^{-1}\left[\frac{2 \sqrt{2}+2}{-2}\right]=\cot ^{-1}(-1-\sqrt{2})=157.5^{\circ}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]=$

Options:
A $\pi-x$
B $2 \pi-x$
C $\frac{x}{2}$
D $\pi-\frac{x}{2}$

The correct answer is:
$\pi-\frac{x}{2}$

Solution:
$\begin{aligned} & \cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right] \\ & =\cot ^{-1}\left[\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}-\sqrt{1+\sin x})} \cdot \frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}\right] \\ & =\cot ^{-1}\left[\frac{(1-\sin x)+(1+\sin x)+2 \sqrt{1-\sin ^2 x}}{(1-\sin x)-(1+\sin x)}\right]\end{aligned}$
$\begin{aligned}= & \cot ^{-1}\left[\frac{2(1+\cos x)}{-2 \sin x}\right]=\cot ^{-1}\left[-\frac{2 \cos ^2(x / 2)}{2 \sin (x / 2) \cos (x / 2)}\right] \\ = & \cot ^{-1}\left(-\cot \frac{x}{2}\right)=\cot ^{-1}\left[\cot \left(\pi-\frac{x}{2}\right)\right]=\pi-\frac{x}{2} .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cot ^{-1} 3+\operatorname{cosec}^{-1} \sqrt{5}=$

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{4}$
C $\frac{\pi}{6}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{4}$

Solution:
$\begin{aligned} & \cot ^{-1} 3+\operatorname{cosec}^{-1} \sqrt{5}=\cot ^{-1} 3+\cot ^{-1} 2 \\ &=\cot ^{-1}\left(\frac{3 \times 2-1}{3+2}\right)=\cot ^{-1}(1)=\frac{\pi}{4}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cot ^{-1}(\sqrt{\cos \alpha})=\tan ^{-1}(\sqrt{\cos \alpha})=x$, then $\sin x=$

Options:
A
$\tan ^2\left(\frac{\alpha}{2}\right)$

B
$\cot ^2\left(\frac{\alpha}{2}\right)$

C
$\tan \alpha$

D
$\cot \left(\frac{\alpha}{2}\right)$

The correct answer is:
$\tan ^2\left(\frac{\alpha}{2}\right)$

Solution:
$\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$
$\tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \Rightarrow \tan ^{-1} \frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}=x$
$\Rightarrow \tan ^{-1} \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=x \Rightarrow \tan x=\frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}$
OR $\cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha}$ or $\operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha}$
$\sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{1-\left(1-2 \sin ^2 \alpha / 2\right)}{1+2 \cos ^2 \alpha / 2-1}$ or $\sin x=\tan ^2 \frac{\alpha}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2002
$\cot ^{-1} x+\tan ^{-1} 3=\frac{\pi}{2}$, then $x=$

Options:
A $1 / 3$
B $1 / 4$
C $3$
D $4$

The correct answer is:
$3$

Solution:
We have
$\cot ^{-1} x+\tan ^{-1} 3=\frac{\pi}{2}$
$\begin{aligned}
& \Rightarrow \cot ^{-1} x+\tan ^{-1} 3=\frac{\pi}{2} \Rightarrow \tan ^{-1} \frac{1}{x}+\tan ^{-1} 3=\frac{\pi}{2} \\
& \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{x}+3}{1-\frac{1}{x} \cdot 3}\right)=\tan ^{-1}\left(\frac{1}{0}\right) \\
& \Rightarrow \frac{3 x+1}{x-3}=\frac{1}{0} \Rightarrow x=3
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cot \left[\cos ^{-1}\left(\frac{7}{25}\right)\right]=$

Options:
A $\frac{25}{24}$
B $\frac{25}{7}$
C $\frac{24}{25}$
D None of these

The correct answer is:
None of these

Solution:
$\cot \left[\cos ^{-1}\left(\frac{7}{25}\right)\right]=\cot \left[\cot ^{-1}\left(\frac{7}{24}\right)\right]=\frac{7}{24}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\cot \theta-\tan \theta-2 \tan 2 \theta-4 \tan 4 \theta=$

Options:
A $4 \cot 8 \theta-\tan 6 \theta$
B $\cot 8 \theta+\tan 3 \theta$
C $\cot 86+\cot 6 \theta$
D $8 \cot 86$

The correct answer is:
$8 \cot 86$

Solution:
$$
\begin{aligned}
& \cot \theta-\tan \theta-2 \tan 2 \theta-4 \tan 4 \theta \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}-2 \frac{\sin 2 \theta}{\cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta} \\
& =2 \frac{\cos ^2 \theta-\sin ^2 \theta}{2 \sin \theta \cos \theta}-2 \frac{\sin 2 \theta}{\cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta} \\
& =2 \frac{\cos 2 \theta}{\sin 2 \theta}-\frac{2 \sin 2 \theta}{\cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 2 \cdot 2 \frac{\left[\cos ^2 2 \theta-\sin ^2 2 \theta\right]}{2 \sin 2 \theta \cdot \cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta} \\
& \Rightarrow \frac{4 \cos 4 \theta}{\sin 4 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta}=4\left[\frac{\cos ^2 4 \theta-\sin ^2 4 \theta}{\sin 4 \theta \cos 4 \theta}\right]=8 \frac{\cos 8 \theta}{\sin 8 \theta} \\
& \Rightarrow \quad 8 \cot 8 \theta
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
Cotton fibres mainly contains

Options:
A cellulose
B glycogen
C protein
D lipid.

The correct answer is:
cellulose

Solution:
The cotton fibres represent epidermal prolongations of seed coat cells. A raw cotton fibre, on average, gives the following values on analysis : cellulose, $94 \%$, protein - $1.3 \%$, pectic substances $-0.9 \%$, ash $-1.2 \%$, wax $-0.6 \%$, sugars $-0.3 \%$ and traces of pigments and mineral matter.

Topic: Biology, Morphology of Flowering Plants, AIIMS, AIIMS 2009
Cotyledons and testa respectively are edible parts in -

Options:
A Cashew nut and litchi
B Groundnut and pomegranate
C Walnut and tamarind
D French bean and coconut

The correct answer is:
Groundnut and pomegranate

Solution:
No Solution Available

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2009 (Mains)
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?

Options:

None

Solution:
The number of eggs given by oviparous and young ones given by viviparous animals are not equal because the egg laying animals lay-more eggs to resist the environmental forces for survival so that the population remains constant or not declining while the viviparous animals nurture their young ones and give birth to less number of offsprings as parental care is shown in the viviparous animals.

Topic: Biology, Animal Kingdom
Covalent compounds are generally ...... in water

Options:
A Soluble
B Insoluble
C Dissociated
D Hydrolysed

The correct answer is:
Insoluble

Solution:
Water is a polar solvent while covalent compounds are non-polar so they usually insoluble in water.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Covalent compounds have low melting point because

Options:
A Covalent bond is less exothermic
B Covalent molecules have definite shape
C Covalent bond is weaker than ionic bond
D Covalent molecules are held by weak Vander Waal's Force

The correct answer is:
Covalent molecules are held by weak Vander Waal's Force

Solution:
The melting and the boiling points of covalent compounds are generally low because of the weak Van der Waals forces and so can be easily converted from liquid to gas or from solid to liquid.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Cow milk, an example of natural emulsion, is stabilised by

Options:
A fat
B water
C casein
D $\mathrm{Mg}^{2+}$ ions

The correct answer is:
casein

Solution:
Cow milk an example of a neutral emulsion is stabilized by casein.
Cow milk is liquid fat particles dispersed in water. It is oil in water type emulsion. Casein is protective colloid with gold number \(0.01-0.02\).
Casein is a phosphoprotein which occurs in the milk of mammals as a suspension of calcium caseinate. It is the main protein of milk.

Topic: Chemistry, Surface Chemistry, JIPMER, JIPMER 2012
Critical angle is that angle of incidence in the denser medium for which the angle of refraction in rarer medium is

Options:
A $0^{\circ}$
B $57^{\circ}$
C $90^{\circ}$
D $180^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
For an angle $i_c$ called critical angle the three possible cases are
1. When $i \lt i_c$, then the incident refracts and the angle it makes with the normal is called the angle of refraction.
2. When $i=i_c$, then the incident ray refracting and coincides with the refracting surface. The angle of refraction is $90^{\circ}$ in this case.
3. When $i\gt i_c$, then the incident ray is totally reflected back and this incident

Topic: Physics, Ray Optics, JEE Main
Critical angle of incidence for a pair of optical media is $45^{\circ}$. The refractive indices of first and second media are in the ratio:

Options:
A $1: \sqrt{2}$
B $\sqrt{2}: 1$
C $2: 1$
D $1: 2$

The correct answer is:
$\sqrt{2}: 1$

Solution:
$\begin{aligned} & \sin \theta_{\mathrm{c}}=\frac{\mu_{\mathrm{R}}}{\mu_{\mathrm{d}}}=\frac{\mu_2}{\mu_1} \\ & \sin 45^{\circ}=\frac{\mu_2}{\mu_1} \\ & \Rightarrow \frac{1}{\sqrt{2}}=\frac{\mu_2}{\mu_1} \\ & \Rightarrow \frac{\mu_1}{\mu_2}=\frac{\sqrt{2}}{1}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Critical angle of light passing from glass to air is minimum for

Options:
A Red
B Green
C Yellow
D Violet

The correct answer is:
Violet

Solution:
${ }_a \mu_g=\frac{1}{\sin C} \Rightarrow \sin C=\frac{1}{{ }_a \mu_g}$
As $\mu$ for violet colour is maximum, so $\sin C$ is minimum and hence critical angle $C$ is minimum for voilet colour.

Topic: Physics, Ray Optics, JEE Main
Crocodile and penguin are similar to whale and dogfish in which one of the following features?

Options:
A possess a solid single-stranded central nervous system
B lay eggs and guard them till they hatch
C possess bony skeleton
D have gill slits at some stage

The correct answer is:
have gill slits at some stage

Solution:
Crocodile, penguin, whale and dogfish, all are chordates. So, all of these have gill slits at some stage of development.

Topic: Biology, Evolution, NEET, NEET 2010 (Mains)
Cross - sectional area of a Copper wire is equal to area of a square of length \( 2 \mathrm{~mm} \). If this copper wire draws \( 8 \) A electric current, then find the drift velocity of free electron. Number density of electron in Copper wire is \( 8 \times 10^{28} \mathrm{~m}^{-3} \).

Options:
A \( 1.56 \times 10^{-4} \mathrm{~ms}^{-1} \)
B \( 1.56 \times 10^{-2} \mathrm{~ms}^{-1} \)
C \( 3.12 \times 10^{-3} \mathrm{~ms}^{-1} \)
D \( 3.12 \times 10^{-2} \mathrm{~ms}^{-1} \)

The correct answer is:
\( 1.56 \times 10^{-4} \mathrm{~ms}^{-1} \)

Solution:
For the given conductor having a square cross-section of sides of 2 mm, area A=2 mm×2 mm
=4 mm2
=4×10-32 m2
⇒A=4×10-6 m2

When the number density of free electrons n, drift velocity Vd and charge of electron e is known, then the current
i=ne AVd ...(1)

Substituting the given values in equation (1),
8×1028×1.6×10-19×4×10-6 Vd=8
⇒Vd=88×1028×1.6×10-19×4×10-6
=1.56×10-4 m s-1

Topic: Physics, Current Electricity, JEE Main
Cross-section view of a prism is the equilateral triangle ABC shown in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ___________ ×10-10 s. (Given, speed of light in vacuum =3×108 m s-1 and cos30°=32)

Options:

The correct answer is:
5

Solution:
δmin=2i-A
A=2i-A
A=i
μ=sinδmin+A2sinA2=sin(A)sinA2=sin60°sin30°
μ=3
v=cμ=3×1083 m s-1
t=APv=53×10-23×1083=5×10-10s
So, x=5

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Cry 1 endotoxins obtained from Bacillus Thuringiensis are effective against

Options:
A Boll worms
B Mosquitoes
C Flies
D Nematodes

The correct answer is:
Boll worms

Solution:
Proteins encoded by the genes cry I Ac and cry II Ab control cotton bolloworm.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Mains)
Cry-I endotoxins obtained from Bacillus thuringiensis are effective against

Options:
A mosquitoes
B flies
C nematodes
D boll worms

The correct answer is:
boll worms

Solution:
A bollworm is a common term for any larva of a moth that attacks the fruiting bodies of certain crops, especially cotton. B. thuringiensis (commonly known as Bt ) is an insecticidal bacterium, marketed worldwide for control of many important plant pests - mainly caterpillars of the Lepidoptera (butterflies and moths) but also mosquito larvae. During sporulation, many Bt strains produce crystal proteins (proteinaceous inclusions), called as endotoxins or cry proteins, that have insecticidal action. This has led to their use as insecticides, and more recently to genetically modified crops using Bt genes.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Screening)
Cryolite is

Options:
A Magnesium sulphate
B Sodium borofluoride
C Sodium aluminium fluoride
D Magnesium silicate

The correct answer is:
Sodium aluminium fluoride

Solution:
Na3AlF6 Sodium hexafluoro aluminate (III)
Cryolite occurs in white masses,and it is used as a flux in the electrolytic production of aluminum.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Crystal field splitting energies for octahedral $\left(\Delta_0\right)$ and tetrahedral $\left(\Delta_t\right)$ geometries caused by the same ligands are related through the expression

Options:
A $\Delta_0=\Delta_t$
B $4 \Delta_0=9 \Delta_1$
C $9 \Delta_0=4 \Delta_1$
D $\Delta_0=2 \Delta_1$

The correct answer is:
$4 \Delta_0=9 \Delta_1$

Solution:
Octahedral splitting energy $\left(\Delta_0\right)$ and tetrahedral splitting energy $\left(\Delta_t\right)$.
For same ligands
$$
\begin{aligned}
\Delta_t & =\frac{4}{9} \Delta_0 \\
9 \Delta_t & =4 \Delta_0
\end{aligned}
$$

Topic: Chemistry, Coordination Compounds, JEE Main
Crystal field splitting energies for octahedral \( \left(\Delta_{0}\right) \) and tetrahedral \( \left(\Delta_{t}\right) \) geometries caused by the same ligands are related through the expression

Options:
A \( \Delta_{0}=\Delta_{\mathrm{t}} \)
B \( 4 \Delta_{0}=9 \Delta_{\mathrm{t}} \)
C \( 9 \Delta_{0}=4 \Delta_{\mathrm{t}} \)
D \( \Delta_{0}=2 \Delta_{\mathrm{t}} \)

The correct answer is:
\( 4 \Delta_{0}=9 \Delta_{\mathrm{t}} \)

Solution:
The relation between octahedral splitting energy (donated by Δ0) and tetrahedral splitting energy (denoted by Δt) is as follows:

Δt=49Δ0

9Δt=4Δ0

Topic: Chemistry, Coordination Compounds, JEE Main
Crystal field stabilization energy for high spin d4 octahedral complex is

Options:
A â0.6â Î 0
B −1.8  Δ 0
C â1.6â Î 0 +âP
D −1.2  Δ 0

The correct answer is:
−0.6  Δ 0

Solution:
For high spin d4 Octahedral complex the splitting is given by
CFSE =  −0.4−× 3 + 0.6 × 1 Δ 0
= −1.2+0.6 Δ 0
=−0.6  Δ 0

Topic: Chemistry, Coordination Compounds, JEE Main
CsBr has bcc structure with edge length 4.3 Å . The shortest inter ionic distance in between Cs + and B r − is –

Options:
A 4.3 Ã
B 7.44 &angst;
C 1.86 &angst;
D 3.72 Ã

The correct answer is:
3.72 Ã

Solution:
In BCC structure the cations and anions touch along the body diagonal
2 r cs + + r Br - = 3 Ã 4 Â· 3 (Body Diagonal = 3 a )
r cs + + r Br - = 1 Â· 7 3 2 Ã 4 Â· 3 2 = 3 Â· 7 2 Ã

Topic: Chemistry, Solid State, JEE Main
CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expressions is correct?

Options:
A r Cs + + r Cl - = 3 a
B r Cs + + r Cl - = 3 a 2
C r Cs + + r Cl - = 3 2 a
D r Cs + + r Cl - = 3 a

The correct answer is:
r Cs + + r Cl - = 3 2 a

Solution:
8 chloride ions at 8 corners Cs + cation at body centre along body diagonal there will be one Cs + cation and 2  Cl − ions at ends of body diagonal.
2 r Cs + + 2 r Cl - = 3 a
a is edge length

Topic: Chemistry, Solid State, JEE Main, JEE Main 2014 (06 Apr)
CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expressions is correct?

Options:
A r Cs + + r Cl - = 3 a
B r Cs + + r Cl - = 3 a 2
C r Cs + + r Cl - = 3 2 a
D r Cs + + r Cl - = 3 a

The correct answer is:
r Cs + + r Cl - = 3 2 a

Solution:
8 chloride ions at 8 corners Cs + cation at body centre along body diagonal there will be one Cs + cation and 2â Cl â ions at ends of body diagonal.
2 r Cs + + 2 r Cl - = 3 a
a is edge length

Topic: Chemistry, Solid State, JEE Main
Cu 2+  +  2e − →Cu; log [ Cu 2+ ] vs.  E red graph is of the type as shown in figure where OA=0.34 V then electrode potential of the half cell of Cu|Cu2+(0.1M) will be

Options:
A -0.34+0.05912 V
B 0.34+0.0591 V
C 0.34 V
D -0.34 V

The correct answer is:
-0.34+0.05912 V

Solution:
ECu/Cu2+=ECu/Cu2+0-0.0592 logCu2+

If logCu2+=0 i.e., Cu2+=1, then ECu/Cu2+=ECu/Cu2+o

or ECu/Cu2+o=-ECu2+/Cuo=-0.34

Now, ECu/Cu2+=-0.34-0.0592log 0.1

=-0.34+0.0592

Topic: Chemistry, Electrochemistry, JEE Main
Cu2+ salt reacts with potassium iodide to give

Options:
A Cu2I2
B Cu2I3
C Cul
D CuI32

The correct answer is:
Cu2I2

Solution:
2Cu+2+4I-⟶Cu2I2s+I2

2Cu+2+3I-⟶2CuI+I2

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (20 Jul Shift 2)