Mischmetal is an alloy consisting mainly of:

Question: Mischmetal is an alloy consisting mainly of: lanthanoid metalsactinoid and transition metalslanthanoid and actinoid metalsactinoid metalsCorrect Option: 1 Solution: Mischmetal is an alloy consisting mainly of lanthanoid metals. Alloy $\Rightarrow$ Lan. metal $\Rightarrow \approx 95 \%$ Iron $\Rightarrow \approx 5 \%$ $\mathrm{~S}, \mathrm{C}, \mathrm{Ca}, \mathrm{Al} \Rightarrow$ traces...

Read More →

The lanthanoid that does NOT show +4 oxidation state is:

Question: The lanthanoid that does NOT show $+4$ oxidation state is:DyCeEuTbCorrect Option: , 3 Solution: Europium(Eu) Atomic No. $-63$ Electronic configuration - $[\mathrm{Xe}] 4 f^{7} 6 s^{2}$ It shows only $+2$ and $+3$ oxidation state....

Read More →

The magnetic field associated with a light wave is given at the origin by

Question: The magnetic field associated with a light wave is given at the origin by $\mathrm{B}=\mathrm{B}_{0}\left[\sin \left(3.14 \times 10^{7}\right) \mathrm{ct}+\sin \left(6.28 \times 10^{7}\right) \mathrm{ct}\right] .$ If this light falls on a silver plate having a work function of $4.7 \mathrm{eV}$, what will be the maximum kinetic energy of the photoelectrons? $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$(1) $6.82 \math...

Read More →

The contrapositive of the statement

Question: The contrapositive of the statement "If you will work, you will earn money" is:(1) If you will not earn money, you will not work(2) You will earn money, if you will not work(3) If you will earn money, you will work(4) To earn money, you need to workCorrect Option: 1 Solution: Contrapositive of $\mathrm{p} \rightarrow \mathrm{q}$ is $\sim \mathrm{q} \rightarrow \sim \mathrm{p}$ $\mathbf{p} \rightarrow$ you will work $\mathrm{q} \rightarrow$ you will earn money $\sim q \rightarrow$ you w...

Read More →

The set that contains atomic numbers of only transition elements, is :

Question: The set that contains atomic numbers of only transition elements, is :$37,42,50,64$$21,25,42,72$$9,17,34,38$$21,32,53,64$Correct Option: , 2 Solution: Elements with atomic number $21,25,42$ and 72 belongs to transition metals....

Read More →

Find the values of p for which the quadratic equation

Question: Find the values of $p$ for which the quadratic equation $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$ has real and equal roots. Solution: The given equation is $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$. This is of the form $a x^{2}+b x+c=0$, where $a=2 p+1, b=-(7 p+2)$ and $c=7 p-3$. $\therefore D=b^{2}-4 a c$ $=[-(7 p+2)]^{2}-4 \times(2 p+1) \times(7 p-3)$ $=\left(49 p^{2}+28 p+4\right)-4\left(14 p^{2}+p-3\right)$ $=49 p^{2}+28 p+4-56 p^{2}-4 p+12$ $=-7 p^{2}+24 p+16$ The given equation will have rea...

Read More →

The statement

Question: The statement $\mathrm{A} \rightarrow(\mathrm{B} \rightarrow \mathrm{A})$ is equivalent to:(1) $\mathrm{A} \rightarrow(\mathrm{A} \wedge \mathrm{B})$(2) $\mathrm{A} \rightarrow(\mathrm{A} \vee \mathrm{B})$(3) $\mathrm{A} \rightarrow(\mathrm{A} \rightarrow \mathrm{B})$(4) $\mathrm{A} \rightarrow(\mathrm{A} \leftrightarrow \mathrm{B})$Correct Option: , 2 Solution: $A \rightarrow(B \rightarrow A)$ $\Rightarrow \mathrm{A} \rightarrow(\sim \mathrm{B} \vee \mathrm{A})$ $\Rightarrow \sim A \v...

Read More →

The correct electronic configuration and spin-only magnetic moment

Question: The correct electronic configuration and spin-only magnetic moment $(\mathrm{BM})$ of $\mathrm{Gd}^{3+}(\mathrm{Z}=64)$, respectively, are:$[\mathrm{Xe}] 4 f^{7}$ and $8.9$$[\mathrm{Xe}] 4 f^{7}$ and $7.9$$[\mathrm{Xe}] 5 f^{7}$ and $8.9$$[\mathrm{Xe}] 5 f^{7}$ and $7.9$Correct Option: , 2 Solution: Electronic configuration of ${ }_{64} \mathrm{Gd}=[\mathrm{Xe}] 4 f^{7} 5 d^{1} 6 s^{2}$ ${ }_{64} \mathrm{Gd}^{3+}=[\mathrm{Xe}] 4 f^{7}$ No. of unpaired electron $=7$ $\mu=\sqrt{n(n+2)} \...

Read More →

For the statements p and q, consider the following compound statements:

Question: For the statements $\mathrm{p}$ and $\mathrm{q}$, consider the following compound statements: (a) $(\sim q \wedge(p \rightarrow q)) \rightarrow \sim p$ (b) $((\mathrm{p} \vee \mathrm{q})) \wedge \sim \mathrm{P}) \rightarrow \mathrm{P}$ Then which of the following statements is correct?(1) (a) is a tautology but not (b)(2) (a) and (b) both are not tautologies.(3) (a) and (b) both are tautologies.(4) (b) is a tautology but not (a).Correct Option: , 3 Solution:...

Read More →

The incorrect statement(s) among (1) - (3) is (are):

Question: The incorrect statement(s) among (1) - (3) is (are): (a) $\mathrm{W}(\mathrm{VI})$ is more stable than $\mathrm{Cr}(\mathrm{VI})$. (b) in the presence of $\mathrm{HCl}$, permanganate titrations provide satisfactory results. (c) some lanthanoid oxides can be used as phosophorus.(b) and (c) only(a) and (b) only(b) only(a) onlyCorrect Option: , 3 Solution: (i) $\mathrm{W}(\mathrm{VI})$ is more stable than $\mathrm{Cr}(\mathrm{VI})$ due to smaller size of atoms and also due to lanthanide c...

Read More →

A particle ' P ' is formed due to a completely inelastic collision

Question: Aparticle ' $\mathrm{P}$ ' is formed due to a completely inelastic collision of particles ' $x$ ' and ' $y$ ' having de-Broglie wavelengths ' $\gamma_{x}$ ' and ' $\gamma_{y}$ ' respectively. If $x$ and $y$ were moving in opposite directions, then the de-Broglie wavelength of ' $\mathrm{P}$ ' is:(1) $\frac{\gamma_{x} \gamma_{y}}{\gamma_{x}+\gamma_{y}}$(2) $\frac{\gamma_{x} \gamma_{y}}{\left|\gamma_{x}-\gamma_{y}\right|}$(3) $\gamma_{x}-\gamma_{y}$(4) $\gamma_{x}+\gamma_{y}$Correct Opti...

Read More →

Solve the following

Question: A $20.0 \mathrm{~mL}$ solution containing $0.2 \mathrm{~g}$ impure $\mathrm{H}_{2} \mathrm{O}_{2}$ reacts completely with $0.316 \mathrm{~g}$ of $\mathrm{KMnO}_{4}$ in acid solution. The purity of $\mathrm{H}_{2} \mathrm{O}_{2}$ (in \%) is________________ (mol. wt. of $\mathrm{H}_{2} \mathrm{O}_{2}=34$; mol. wt. of $\mathrm{KMnO}_{4}=1 \overline{58)}$ Solution: (85)...

Read More →

Find the values of k for which the quadratic equation

Question: (i) Find the values of $k$ for which the quadratic equation $(3 k+1) x^{2}+2(k+1) x+1=0$ has real and equal roots. (ii) Find the value of $k$ for which the equation $x^{2}+k(2 x+k-1)+2=0$ has real and equal roots. Solution: (i) The given equation is $(3 k+1) x^{2}+2(k+1) x+1=0$. This is of the form $a x^{2}+b x+c=0$, where $a=3 k+1, b=2(k+1)$ and $c=1$. $\therefore D=b^{2}-4 a c$ $=[2(k+1)]^{2}-4 \times(3 k+1) \times 1$ $=4\left(k^{2}+2 k+1\right)-4(3 k+1)$ $=4 k^{2}+8 k+4-12 k-4$ $=4 ...

Read More →

The incorrect statement is:

Question: The incorrect statement is:Manganate and permanganate ions are tetrahedralIn manganate and permanganate ions, the $\pi$-bonding takes place by overlap of $p$-orbitals of oxygen and $d$ orbitals of manganeseManganate and permanganate ions are paramagneticManganate ion is green in colour and permanganate ion is purple in colourCorrect Option: , 3 Solution: Manganate ion is paramagnetic while permanganate ion is diamagnetic....

Read More →

The negation of the statement

Question: The negation of the statement $\sim p \wedge(p \vee q)$ is :(1) $\sim p \wedge q$(2) $\mathrm{p} \wedge \sim \mathrm{q}$(3) $\sim p \vee q$(4) $\mathrm{p} \vee \sim \mathrm{q}$Correct Option: , 4 Solution: $\therefore \sim p \wedge(p \vee q) \equiv \mathrm{p} \vee \sim \mathrm{q}$...

Read More →

The electric field of light wave is given as

Question: The electric field of light wave is given as $\vec{E}=10^{3} \cos$ $\left(\frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t\right) \hat{x} \frac{N}{C}$ This light falls on a metal plate of work function $2 \mathrm{eV}$. The stopping potential of the photo-electrons is: Given, $\mathrm{E}($ in $\mathrm{eV})=\frac{12375}{\lambda(\text { in } A)}$(1) $2.0 \mathrm{~V}$(2) $0.72 \mathrm{~V}$(3) $0.48 \mathrm{~V}$(4) $2.48 \mathrm{~V}$Correct Option: 3 Solution: (3) Here $\ome...

Read More →

The electric field of light wave is given as

Question: The electric field of light wave is given as $\vec{E}=10^{3} \cos$ $\left(\frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t\right) \hat{x} \frac{N}{C}$ This light falls on a metal plate of work function $2 \mathrm{eV}$. The stopping potential of the photo-electrons is: Given, $E($ in $\mathrm{eV})=\frac{12375}{\lambda(\text { in } \AA)}$(1) $2.0 \mathrm{~V}$(2) $0.72 \mathrm{~V}$(3) $0.48 \mathrm{~V}$(4) $2.48 \mathrm{~V}$Correct Option: 3 Solution: (3) Here $\omega=2 \p...

Read More →

Find the non-zero value of k for which the roots of the quadratic equation

Question: Find the non-zero value of $k$ for which the roots of the quadratic equation $9 x^{2}-3 k x+k=0$ are real and equal. Solution: The given equation is $9 x^{2}-3 k x+k=0$. This is of the form $a x^{2}+b x+c=0$, where $a=9, b=-3 k$ and $c=k$. $\therefore D=b^{2}-4 a c=(-3 k)^{2}-4 \times 9 \times k=9 k^{2}-36 k$ The given equation will have real and equal roots ifD= 0. $\therefore 9 k^{2}-36 k=0$ $\Rightarrow 9 k(k-4)=0$ $\Rightarrow k=0$ or $k-4=0$ $\Rightarrow k=0$ or $k=4$ But,k 0 (Giv...

Read More →

The oxidation states of transition metal atoms in

Question: The oxidation states of transition metal atoms in $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$, $\mathrm{KMnO}_{4}$ and $\mathrm{K}_{2} \mathrm{FeO}_{4}$, respectively, are $x, y$ and $z$. The sum of $x, y$ and $z$ is_____________ Solution: (19)...

Read More →

The statement among the following that is a tautology is:

Question: The statement among the following that is a tautology is:(1) $A \wedge(A \vee B)$(2) $\mathrm{B} \rightarrow[\mathrm{A} \wedge(\mathrm{A} \rightarrow \mathrm{B})]$(3) $A \vee(A \wedge B)$(4) $[\mathrm{A} \wedge(\mathrm{A} \rightarrow \mathrm{B})] \rightarrow \mathrm{B}$Correct Option: , 4 Solution: $A \wedge(\sim A \vee B) \rightarrow B$ $=[(A \wedge \sim A) \vee(A \wedge B)] \rightarrow B$ $=(A \wedge B) \rightarrow B$ $=\sim A \vee \sim B \vee B$ $=t$...

Read More →

Dichromate ion is treated with base,

Question: Dichromate ion is treated with base, the oxidation number of $\mathrm{Cr}$ in the product formed is: Solution: (6) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-}=2 \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}$ $2 \mathrm{CrO}_{7}^{2-}$ $\mathrm{x}+(-2 \times 4)=-2$ $x=6$...

Read More →

For what values of p are the roots of the equation

Question: For what values of $p$ are the roots of the equation $4 x^{2}+p x+3=0$ real and equal? Solution: The given equation is $4 x^{2}+p x+3=0$. This is of the form $a x^{2}+b x+c=0$, where $a=4, b=p$ and $c=3$. $\therefore D=b^{2}-4 a c=p^{2}-4 \times 4 \times 3=p^{2}-48$ The given equation will have real and equal roots ifD= 0. $\therefore p^{2}-48=0$ $\Rightarrow p^{2}=48$ $\Rightarrow p=\pm \sqrt{48}=\pm 4 \sqrt{3}$ Hence, $4 \sqrt{3}$ and $-4 \sqrt{3}$ are the required values of $p$....

Read More →

In the density measurement of a cube,

Question: In the density measurement of a cube, the mass and edge length are measured as $(10.00 \pm 0.10) \mathrm{kg}$ and $(0.10 \pm 0.01)$ $\mathrm{m}$, respectively. The error in the measurement of density is:(1) $0.01 \mathrm{~kg} / \mathrm{m}^{3}$(2) $0.10 \mathrm{~kg} / \mathrm{m}^{3}$(3) $0.31 \mathrm{~kg} / \mathrm{m}^{3}$(4) $0.07 \mathrm{~kg} / \mathrm{m}^{3}$Correct Option: 3 Solution: (3) $d=\frac{M}{V}=\frac{M}{L^{3}}=M L^{-3}$ $\frac{\Delta d}{d}=\frac{\Delta M}{M}+3 \frac{\Delta ...

Read More →

Which one of the following lanthanoids does not form

Question: Which one of the following lanthanoids does not form $\mathrm{MO}_{2}$ ?NdYbDyPrCorrect Option: Solution: (2) Fact...

Read More →

For what value of k are the roots of the quadratic equation

Question: For what value of $k$ are the roots of the quadratic equation $k x(x-2 \sqrt{5})+10=0$ real and equal? Solution: The given equation is $k x(x-2 \sqrt{5})+10=0$ $\Rightarrow k x^{2}-2 \sqrt{5} k x+10=0$ This is of the form $a x^{2}+b x+c=0$, where $a=k, b=-2 \sqrt{5} k$ and $c=10$. $\therefore D=b^{2}-4 a c=(-2 \sqrt{5} k)^{2}-4 \times k \times 10=20 k^{2}-40 k$ The given equation will have real and equal roots ifD= 0. $\therefore 20 k^{2}-40 k=0$ $\Rightarrow 20 k(k-2)=0$ $\Rightarrow ...

Read More →