Which of the following compounds will show retention in configuration
Question: Which of the following compounds will show retention in configuration on nucleophic substitution by $\mathrm{OH}^{-}$ion?Correct Option: , 4 Solution:...
Read More →Solve the following
Question: Let $\alpha$ be a root of the equation $x^{2}+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 1 1 \\ 1 \alpha \alpha^{2} \\ 1 \alpha^{2} \alpha^{4}\end{array}\right]$, then the matrix $A^{31}$ is equal to:(1) $A$(2) $I_{3}$(3) $A^{2}$(4) $A^{3}$Correct Option: , 4 Solution: Solution of $x^{2}+x+1=0$ is $\omega, \omega^{2}$ So, $\alpha=\omega$ and $\omega^{4}=\omega^{3} \cdot \omega=1 \cdot \omega=\omega$ $A^{2}=\frac{1}{3}\left[\begin{array}{ccc}1 1 1 \\ 1 \omega \...
Read More →If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.
Question: If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term. Solution: Letabe the first term anddbe the common difference of the AP. Then, $4 \times a_{4}=18 \times a_{18} \quad$ (Given) $\Rightarrow 4(a+3 d)=18(a+17 d) \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 2(a+3 d)=9(a+17 d)$ $\Rightarrow 2 a+6 d=9 a+153 d$ $\Rightarrow 7 a=-147 d$ $\Rightarrow a=-21 d$ $\Rightarrow a+21 d=0$ $\Rightarrow a+(22-1) d=0$ $\Rightarrow a_{22}=0$ Hence, the 22nd te...
Read More →A galvanometer having a coil resistance
Question: A galvanometer having a coil resistance $100 \Omega$ gives a full scale deflection when a current of $1 \mathrm{~mA}$ is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of $10 \mathrm{~V}$ ?(1) $10 \mathrm{k} \Omega$(2) $8.9 \mathrm{k} \Omega$(3) $7.9 \mathrm{k} \Omega$(4) $9.9 \mathrm{k} \Omega$Correct Option: 4, Solution: (4) Given, Resistance of galvanometer, $G=100 \O...
Read More →Determine the nth term of the AP whose 7th term is −1 and 16th term is 17.
Question: Determine thenth term of the AP whose 7th term is 1 and 16th term is 17. Solution: Letabe the first term anddbe the common difference of the AP. Then, $a_{7}=-1$ $\Rightarrow a+(7-1) d=-1 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+6 d=-1 \quad \ldots$ (1) Also, $a_{16}=17$ $\Rightarrow a+15 d=17 \quad \ldots(2)$ From (1) and (2), we get $-1-6 d+15 d=17$ $\Rightarrow 9 d=17+1=18$ $\Rightarrow d=2$ Puttingd= 2 in (1), we get $a+6 \times 2=-1$ $\Rightarrow a=-1-12=-13$ $\therefore ...
Read More →For the given reaction :
Question: For the given reaction : What is A$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$Correct Option: , 3 Solution:...
Read More →Prove the following
Question: If $\alpha$ and $\beta$ are the roots of the equation $2 x(2 x+1)=1$, then $\beta$ is equal to:(1) $2 \alpha(\alpha+1)$(2) $-2 \alpha(\alpha+1)$(3) $2 \alpha(\alpha-1)$(4) $2 \alpha^{2}$Correct Option: , 2 Solution: (2) Let $\alpha$ and $\beta$ be the roots of the given quadratic equation, $2 x^{2} \cdot 2 x-1=0$...(i) Then, $\alpha+\beta=-\frac{1}{2} \Rightarrow-1=2 \alpha+2 \beta$ and $4 \alpha^{2}+2 \alpha-1=0$$[\because \alpha$ is root of eq. (i) $]$ $\Rightarrow 4 \alpha^{2}+2 \al...
Read More →The figure gives experimentally measured B vs.
Question: The figure gives experimentally measured $B$ vs. $H$ variation in a ferromagnetic material. The retentivity, co-ercivity and saturation, respectively, of the material are:(1) $1.5 \mathrm{~T}, 50 \mathrm{~A} / \mathrm{m}$ and $1.0 \mathrm{~T}$(2) $1.5 \mathrm{~T}, 50 \mathrm{~A} / \mathrm{m}$ and $1.0 \mathrm{~T}$(3) $150 \mathrm{~A} / \mathrm{m}, 1.0 \mathrm{~T}$ and $1.5 \mathrm{~T}$(4) $1.0 \mathrm{~T}, 50 \mathrm{~A} / \mathrm{m}$ and $1.5 \mathrm{~T}$Correct Option: 2, Solution:...
Read More →The 9th term of an AP is −32 and the sum of its 11th and 13th terms is −94.
Question: The 9th term of an AP is 32 and the sum of its 11th and 13th terms is 94. Find the common difference of the AP. Solution: Letabe the first term anddbe the common difference of the AP. Then, $a_{9}=-32$ $\Rightarrow a+(9-1) d=-32 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+8 d=-32 \quad \ldots \ldots(1)$ Now, $a_{11}+a_{13}=-94$ (Given) $\Rightarrow(a+10 d)+(a+12 d)=-94$ $\Rightarrow 2 a+22 d=-94$ $\Rightarrow a+11 d=-47 \quad \ldots(2)$ From (1) and (2), we get $-32-8 d+11 d=-47$...
Read More →For the given reaction :
Question: For the given reaction : What is 'A'?Correct Option: , 4 Solution: It is bezylic substitution reaction...
Read More →Prove the following
Question: If $\alpha$ and $\beta$ be two roots of the equation $x^{2}-64 x+256=0$. Then the value of $\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}$ is:(1) 2(2) 3(3) 1(4) 4Correct Option: 1 Solution: $\because \alpha+\beta=64, \alpha \beta=256$ $\frac{\alpha^{3 / 8}}{\beta^{5 / 8}}+\frac{\beta^{3 / 8}}{\alpha^{5 / 8}}=\frac{\alpha+\beta}{(\alpha \beta)^{5 / 8}}=\frac{64}{\left(2^{8}\right)^{5 / 8}}=\frac{64}{32}=2$...
Read More →An iron rod of volume
Question: An iron rod of volume $10^{-3} \mathrm{~m}^{3}$ and relative permeability 1000 is placed as core in a solenoid with 10 turns $/ \mathrm{cm}$. If a current of $0.5 \mathrm{~A}$ is passed through the solenoid, then the magnetic moment of the rod will be :(1) $50 \times 10^{2} \mathrm{Am}^{2}$(2) $5 \times 10^{2} \mathrm{Am}^{2}$(3) $500 \times 10^{2} \mathrm{Am}^{2}$(4) $0.5 \times 10^{2} \mathrm{Am}^{2}$Correct Option: , 2 Solution: (2) Given, Volume of iron rod, $V=10^{-3} \mathrm{~m}^...
Read More →Prove the following
Question: If $\alpha$ and $\beta$ are the roots of the equation, $7 x^{2}-3 x-2=0$, the the value of $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ is equal to :(1) $\frac{27}{32}$(2) $\frac{1}{24}$(3) $\frac{3}{8}$(4) $\frac{27}{16}$Correct Option: , 4 Solution: Let $\alpha$ and $\beta$ be the roots of the quadratic equation $7 x^{2}-3 x-2=0$ $\therefore \alpha+\beta=\frac{3}{7}, \alpha \beta=\frac{-2}{7}$ Now, $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ $=\frac{\alpha-\alp...
Read More →The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34.
Question: The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference. Solution: Letabe the first term anddbe the common difference of the AP. Then, $a_{4}=11$ $\Rightarrow a+(4-1) d=11 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+3 d=11 \quad \ldots \ldots(1)$ Now, $a_{5}+a_{7}=34$ (Given) $\Rightarrow(a+4 d)+(a+6 d)=34$ $\Rightarrow 2 a+10 d=34$ $\Rightarrow a+5 d=17 \quad \ldots . .(2)$ From (1) and (2), we get $11-3 d+5 d=17$ $\Rightarrow 2...
Read More →Water does not produce CO on reacting with :
Question: Water does not produce $\mathrm{CO}$ on reacting with :$\mathrm{C}_{3} \mathrm{H}_{8}$C$\mathrm{CH}_{4}$$\mathrm{CO}_{2}$Correct Option: , 4 Solution: $\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}$...
Read More →A paramagnetic sample shows a net magnetisation
Question: A paramagnetic sample shows a net magnetisation of $6 \mathrm{~A} / \mathrm{m}$ when it is placed in an external magnetic field of $0.4 \mathrm{~T}$ at a temperature of $4 \mathrm{~K}$. When the sample is placed in an external magnetic field of $0.3 \mathrm{~T}$ at a temperature of $24 \mathrm{~K}$, then the magnetisation will be:(1) $1 \mathrm{~A} / \mathrm{m}$(2) $4 \mathrm{~A} / \mathrm{m}$(3) $2.25 \mathrm{~A} / \mathrm{m}$(4) $0.75 \mathrm{~A} / \mathrm{m}$Correct Option: , 4 Solu...
Read More →Consider the following chemical reaction.
Question: Consider the following chemical reaction. The number of $s p^{2}$ hybridized carbon atom(s) present in the product is Solution: (7) All carbon atoms in benzaldehyde are $\mathrm{sp}^{2}$ hybridised...
Read More →The product of the roots of the equation
Question: The product of the roots of the equation $9 x^{2}-18|x|+5=$ 0 , is :(1) $\frac{5}{9}$(2) $\frac{25}{81}$(3) $\frac{5}{27}$(4) $\frac{25}{9}$Correct Option: , 2 Solution: Let $|x|=y$ then $9 y^{2}-18 y+5=0$ $\Rightarrow 9 y^{2}-15 y-3 y+5=0$ $\Rightarrow(3 y-1)(3 y-5)=0$ $\Rightarrow y=\frac{1}{3}$ or $\frac{5}{3} \Rightarrow|x|=\frac{1}{3}$ or $\frac{5}{3}$ Roots are $\pm \frac{1}{3}$ and $\pm \frac{5}{3}$ $\therefore$ Product $=\frac{25}{81}$...
Read More →If the sixth term of an AP is zero then show that its 33rd term is three times its 15th term.
Question: If the sixth term of an AP is zero then show that its 33rd term is three times its 15th term. Solution: $a_{6}=a+5 d=0$ And $a_{33}=a+32 d=a+5 d+27 d$ $=27 d \quad(\because a+5 d=0) \quad \ldots(1)$ 3 times of 15 th term means $3 a_{15}=3(a+14 d)=3 a+42 d$ $=3 a+15 d+27 d$ $=3(a+5 d)+27 d$ $=27 d$ $\cdots \cdot(2)$ Thus, (1) = (2)Hence, 33rd term is three times its 15th term....
Read More →A perfectly diamagnetic sphere has a small spherical cavity at its centre,
Question: A perfectly diamagnetic sphere has a small spherical cavity at its centre, which is filled with a paramagnetic substance. The whole system is placed in a uniform magnetic field $\vec{B}$. Then the field inside the paramagnetic substance is : (1) $\vec{B}$(2) zero(3) much large than $|\vec{B}|$ and parallel to $\vec{B}$(4) much large than $|\vec{B}|$ but opposite to $\vec{B}$Correct Option: , 2 Solution: (2) When magnetic field is applied to a diamagnetic substance, it produces magnetic...
Read More →Prove the following
Question: Let $\lambda \neq 0$ be in $\mathbf{R}$. If $\alpha$ and $\beta$ are roots of the equation, $x^{2}-x+2 \lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3 x^{2}-10 x+27 \lambda=0$, then $\frac{\beta \gamma}{\lambda}$ is equal to :(1) 27(2) 18(3) 9(4) 36Correct Option: , 2 Solution: Since $\alpha$ is common root of $x^{2}-x+2 \lambda=0$ and $3 x^{2}-10 x+27 \lambda=0$ $\therefore 3 \alpha^{2}-10 \alpha+27 \lambda=0$...(i) $3 \alpha^{2}-3 \alpha+6 \lambda=0$..(ii) $\th...
Read More →Identify A in the given chemical reaction.
Question: Identify A in the given chemical reaction. Correct Option: , 4 Solution: Aromatization reaction or hydroforming reaction....
Read More →The 4th term of an AP is zero. Prove that its 25 term is triple is 11th term.
Question: The 4th term of an AP is zero. Prove that its 25 term is triple is 11th term. Solution: In the given AP, let the first term beaand the common difference bed.Then, Tn=a+ (n- 1)dNow, T4=a+ (4- 1)d⇒a+3d = 0 ...(1) ⇒ a = -3d Again, T11=a+ (11 - 1)d =a+10d= -3d+ 10d= 7d [ Using (1)]Also, T25=a+ (25 - 1)d =a+24d= -3d+ 24d= 21d [ Using (1)] i.e., T25= 3 ⨯ 7d= (3 ⨯ T11)Hence, 25thterm is triple its 11thterm....
Read More →Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties of the following,
Question: Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties of the following, which property best matches for the type of magnet required?(1) $\mathrm{T}$ : Large retentivity, small coercivity(2) $\mathrm{P}$ : Small retentivity, large coercivity(3) $T$ : Large retentivity, large coercivity(4) P : Large retentivity, large coercivityCorrect Option: , 4 Solution: (4) Permanent magnets $(P)$ are made of materials with large retentivi...
Read More →The 7th term of an AP is −4 and its 13th term is −6. Find the AP.
Question: The 7th term of an AP is 4 and its 13th term is 6. Find the AP. Solution: We have:T7=a+ (n- 1)d⇒a+6d= -4 ...(1)T13=a+ (n- 1)d⇒a+12d= -16 ...(2)On solving (1) and (2), we get:a= 8 andd= -2 Thus, first term = 8 and common difference = -2 The terms of the AP are 8, 6, 4, 2,......
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