Find how many integers between 200 and 500 are divisible by 8.
Question: Find how many integers between 200 and 500 are divisible by 8. Solution: Numbers between 200 and 500 divisible by 8 are 208, 216, ..., 496.This forms an AP 208, 216, ..., 496.So, first term (a) = 208Common difference (d) = 8 $a_{n}=a+(n-1) d=496$ $\Rightarrow 208+(n-1) 8=496$ $\Rightarrow(n-1) 8=288$ $\Rightarrow n-1=36$ $\Rightarrow n=37$ Thus, there are 37 integers between 200 and 500 which are divisible by 8....
Read More →Prove the following
Question: If $\alpha, \beta$ are natural numbers such that $100^{\alpha}-199 \beta=(100)(100)+(99)(101)+(98)(102)+\ldots \ldots+(1)(199)$ then the slope of the line passing through $(\alpha, \beta)$ and origin is :(1) 540(2) 550(3) 530(4) 510Correct Option: 2, Solution: $S=(100)(100)+(99)(101)+(98)(102) \ldots \ldots .(2)(198)+(1)(199)$ $S=\sum_{x=0}^{99}(100-x)(100+x)=\sum 100^{2}-x^{2}$ $=100^{3}-\frac{99 \times 100 \times 199}{6}$ $\alpha=3 \quad \beta=1650$ slope $=\frac{1650}{3}=550$...
Read More →A sum of ₹2800 is to be used to award four prizes.
Question: A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less than the preceding prize, find the value of each of the prizes. Solution: Let the amount of the first prize be ₹a.Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the four prizes are in AP.Amount of the second prize = ₹(a 200)Amount of the third prize = ₹(a 2 200) = ₹(a 400)Amount of the fourth prize = ₹(a 3 200) = ₹(a 600)Now,Total sum of the four priz...
Read More →Consider a solid sphere of radius $R$ and mass density
Question: Consider a solid sphere of radius $R$ and mass density $\rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right), 0r \leq R .$ The minimum density of a liquid in which it will float is:(1) $\frac{\rho_{0}}{3}$(2) $\frac{\rho_{0}}{5}$(3) $\frac{2 \rho_{0}}{5}$(4) $\frac{2 \rho_{0}}{3}$Correct Option: , 3 Solution: (3) For minimum density of liquid, solid sphere has to float (completely immersed) in the liquid. $m g=F_{B}\left(\right.$ also $\left.V_{\text {immersed }}=V_{\text {total }}\right...
Read More →The major product of the following reaction is :
Question: The major product of the following reaction is : Correct Option: Solution:...
Read More →In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on.
Question: In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed? Solution: The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11. Difference of rose plants between two consecutive rows = (41 - 43) = (39 - 41) =-2 [Constant]So, the given progression is an AP.Here, first term = 43Common difference = -2Last term = 11Letnbe the last term, then we hav...
Read More →Prove the following
Question: If $1, \log _{10}\left(4^{x}-2\right)$ and $\log _{10}\left(4^{x}+\frac{18}{5}\right)$ are in arithmetic progression for a real number $x$, then the value of the determinant $\left|\begin{array}{ccc}2\left(\begin{array}{cc}x-\frac{1}{2} \\ 1\end{array}\right. x-1 x^{2} \\ x 1 0\end{array}\right|$ is equal to : Solution: $2 \log _{10}\left(4^{x}-2\right)=1+\log _{10}\left(4^{x}+\frac{18}{5}\right)$ $\left(4^{x}-2\right)^{2}=10\left(4^{x}+\frac{18}{5}\right)$ $\left(4^{x}\right)^{2}+4-4\...
Read More →How many numbers are there between 101 and 999, which are divisible by both 2 and 5?
Question: How many numbers are there between 101 and 999, which are divisible by both 2 and 5? Solution: The numbers which are divisible by both 2 and 5 are divisible by 10 also.Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130, ..., 990.Clearly, these number are in AP.Here,a= 110 andd= 120 110 = 10Let this AP containsnterms. Then, $a_{n}=990$ $\Rightarrow 110+(n-1) \times 10=990 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 10 n+100=990$ $\Rightarrow 10 n=990-100=8...
Read More →But 2-ene on reaction with alkaline
Question: But 2-ene on reaction with alkaline $\mathrm{KMnO}_{4}$ at elevated temperature followed by acidification will give :2 molecules of $\mathrm{CH}_{3} \mathrm{CHO}$Correct Option: , 3 Solution:...
Read More →The sum of two forces
Question: The sum of two forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ such that $|\vec{R}|=|\vec{P}|$. The angle $\theta$ (in degrees) that the resultant of 2 $\vec{P}$ and $\vec{Q}$ will make with $\vec{Q}$ is__________ Solution: (90) Given, $|\vec{R}|=|\vec{P}| \Rightarrow|\vec{P}+\vec{Q}|=|\vec{P}|$ $P^{2}+Q^{2}+2 P Q \cdot \cos \theta=P^{2}$ $\Rightarrow Q+2 P \cos \theta=0$ $\Rightarrow \cos \theta=-\frac{Q}{2 P}$ ...(i) $\tan \alpha=\frac{2 P \sin \theta}{Q+2 P \cos \theta}=\infty(\because ...
Read More →How many three-digit numbers are divisible by 9?
Question: How many three-digit numbers are divisible by 9? Solution: The three-digit numbers divisible by 9 are 108, 117, 126, ..., 999.Clearly, these number are in AP.Here,a= 108 andd= 117 108 = 9Let this AP containsnterms. Then, $a_{n}=999$ $\Rightarrow 108+(n-1) \times 9=999 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 9 n+99=999$ $\Rightarrow 9 n=999-99=900$ $\Rightarrow n=100$ Hence, there are 100 three-digit numbers divisible by 9....
Read More →The major product of the following addition reaction is
Question: The major product of the following addition reaction is Correct Option: Solution:...
Read More →A student measuring the diameter of a pencil of circular cross-section
Question: A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings $5.50 \mathrm{~mm}, 5.55 \mathrm{~mm}, 5.45 \mathrm{~mm}$, $5.65 \mathrm{~mm}$, The average of these four reading is $5.5375 \mathrm{~mm}$ and the standard deviation of the data is $0.07395 \mathrm{~mm}$. The average diameter of the pencil should therefore be recorded as:(1) $(5.5375 \pm 0.0739) \mathrm{mm}$(2) $(5.5375 \pm 0.0740) \mathrm{mm}$(3) ...
Read More →Prove the following
Question: Let $S_{n}(x)=\log _{a^{1 / 2}} x+\log _{a^{1 / 3}} x+\log _{a^{1 / 6}} x$ $+\log _{a^{1 / 1}} x+\log _{a^{1 / 18}} x+\log _{a^{1 / 27}} x+\ldots$ up to $\mathrm{n}$-terms, where $\mathrm{a}1$. If $\mathrm{S}_{24}(\mathrm{x})=1093$ and $\mathrm{S}_{12}(2 \mathrm{x})=265$, then value of a is equal to_______. Solution: $\mathrm{S}_{\mathrm{n}}(\mathrm{x})=(2+3+6+11+18+27+\ldots \ldots+\mathrm{n}-$ terms $) \log _{\mathrm{a}} \mathrm{x}$ Let $\mathrm{S}_{1}=2+3+6+11+18+27+\ldots+\mathrm{T...
Read More →How many two-digit numbers are divisible by 3?
Question: How many two-digit numbers are divisible by 3? Solution: The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.Clearly, these number are in AP.Here,a= 12 andd= 15 12 = 3Let this AP containsnterms. Then, $a_{n}=99$ $\Rightarrow 12+(n-1) \times 3=99 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 3 n+9=99$ $\Rightarrow 3 n=99-9=90$ $\Rightarrow n=30$ Hence, there are 30 two-digit numbers divisible by 3....
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: $\mathrm{CH}_{3} \mathrm{CD}(\mathrm{I}) \mathrm{CHD}(\mathrm{Cl})$$\mathrm{CH}_{3} \mathrm{CD}(\mathrm{Cl}) \mathrm{CHD}$ (I)$\mathrm{CH}_{3} \mathrm{CD}_{2} \mathrm{CH}(\mathrm{Cl})$ (I)$\mathrm{CH}_{3} \mathrm{C}(\mathrm{I})(\mathrm{Cl}) \mathrm{CHD}_{2}$Correct Option: , 4 Solution:...
Read More →The density of a solid metal sphere is determined by measuring its mass and its diameter.
Question: The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right) \%$. If the relative errors in measuring the mass and the diameter are $6.0 \%$ and $1.5 \%$ respectively, the value of $x$ is________ Solution: $(1050)$ Density, $\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3}} \Rightarrow \rho=\frac{6}{\pi} M D^{-3}$ $\therefore \%\left(\frac{\Delta \rho}{\rho}\r...
Read More →Prove the following
Question: Let $\frac{1}{16}$, a and $b$ be in G.P. and $\frac{1}{a}, \frac{1}{b}, 6$ be in A.P., where $a, b0$. Then $72(a+b)$ is equal to__________. Solution: $a^{2}=\frac{b}{16} \Rightarrow \frac{1}{b}=\frac{1}{16 a^{2}}$ $\frac{2}{b}=\frac{1}{a}+6$ $\frac{1}{8 a^{2}}=\frac{1}{a}+6$ $\frac{1}{a^{2}}-\frac{8}{a}-48=0$ $\frac{1}{a}=12,-4 \Rightarrow a=\frac{1}{12},-\frac{1}{4}$ $a=\frac{1}{12}, a0$ $b=16 a^{2}=\frac{1}{9}$ $\Rightarrow \quad 72(a+b)=6+8=14$...
Read More →A physical quantity z depends on four observables a, b, c
Question: A physical quantity $z$ depends on four observables $a, b, c$ and $d$, as $z=\frac{a^{2} b^{\frac{2}{3}}}{\sqrt{c} d^{3}} .$ The percentages of error in the measurement of $a, b, c$ and $d$ are $2 \%, 1.5 \%, 4 \%$ and $2.5 \%$ respectively. The percentage of error in $z$ is :(1) $12.25 \%$(2) $16.5 \%$(3) $13.5 \%$(4) $14.5 \%$Correct Option: , 4 Solution: (4) Given : $Z=\frac{a^{2} b^{2 / 3}}{\sqrt{c} d^{3}}$ Percentage error in $Z$, $=\frac{\Delta Z}{Z}=\frac{2 \Delta a}{a}+\frac{2}...
Read More →Which one of the following alkenes when treated with
Question: Which one of the following alkenes when treated with $\mathrm{HCl}$ yields majorly an anti Markovnikov product?$\mathrm{CH}_{3} \mathrm{O}-\mathrm{CH}=\mathrm{CH}_{2}$$\mathrm{Cl}-\mathrm{CH}=\mathrm{CH}_{2}$$\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}=\mathrm{CH}_{2}$$\mathrm{F}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}$Correct Option: , 4 Solution:...
Read More →Which one of the following alkenes when treated with
Question: Which one of the following alkenes when treated with $\mathrm{HCl}$ yields majorly an anti Markovnikov product?$\mathrm{CH}_{3} \mathrm{O}-\mathrm{CH}=\mathrm{CH}_{2}$$\mathrm{Cl}-\mathrm{CH}=\mathrm{CH}_{2}$$\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}=\mathrm{CH}_{2}$$\mathrm{F}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}$Correct Option: Solution:...
Read More →Consider an arithmetic series and a geometric
Question: Consider an arithmetic series and a geometric series having four initial terms from the set $\{11,8,21,16,26,32,4\}$. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to Solution: GP : $4,8,16,32,64,128,256,512,1024,2048$ 4096,8192 $\mathrm{AP}: 11,16,21,26,31,36$ Common terms: $16,256,4096$ only...
Read More →A balloon is moving up in air vertically above a point
Question: A balloon is moving up in air vertically above a point $\mathrm{A}$ on the ground. When it is at a height $h_{1}$, a girl standing at a distance $d$ (point B) from A (see figure) sees it at an angle $45^{\circ}$ with respect to the vertical. When the balloon climbs up a further height $h_{2}$, it is seen at an angle $60^{\circ}$ with respect to the vertical if the girl moves further by a distance $2.464 d$ (point C). Then the height $h_{2}$ is (given $\tan 30^{\circ}=$ $0.5774)$ : (1) ...
Read More →A balloon is moving up in air vertically above a point
Question: A balloon is moving up in air vertically above a point $\mathrm{A}$ on the ground. When it is at a height $h_{1}$, a girl standing at a distance $d$ (point B) from A (see figure) sees it at an angle $45^{\circ}$ with respect to the vertical. When the balloon climbs up a further height $h_{2}$, it is seen at an angle $60^{\circ}$ with respect to the vertical if the girl moves further by a distance $2.464 d$ (point C). Then the height $h_{2}$ is (given $\tan 30^{\circ}=$ $0.5774)$ : (1) ...
Read More →The correct order of heat of combustion for following alkadienes is:
Question: The correct order of heat of combustion for following alkadienes is: $(\mathrm{A})(\mathrm{B})(\mathrm{C})$(A) $(\mathrm{C})(\mathrm{B})$$(\mathrm{C})(\mathrm{B})(\mathrm{A})$$(\mathrm{B})(\mathrm{C})(\mathrm{A})$Correct Option: 1 Solution: In isomers of hydrocarbon heat of combustion is inversely proportional to the stability. Stability order: $ABC$ Order of heat of combustion: $ABC$...
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