Solve the following
Question: Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is :(1) $3: 1$(2) $9: 7$(3) $5: 3$(4) $33: 31$Correct Option: , 2 Solution: Let $\alpha, \beta, \gamma, \delta$ be in G.P., then $\alpha \delta=\beta \gamma$ $\Rightarrow \frac{\alpha}{\beta}=\frac{\gamma}{\delta} \Rightarrow\left|\frac{\alpha-\beta}{\alpha+\beta}\right|=\left|\f...
Read More →In a ferromagnetic material, below the curie temperature, a domain is defined as:
Question: In a ferromagnetic material, below the curie temperature, a domain is defined as:(1) a macroscopic region with consecutive magnetic diploes oriented in opposite direction.(2) a macroscopic region with zero magnetization.(3) a macroscopic region with saturation magnetization.(4) a macroscopic region with randomly oriented magnetic dipoles.Correct Option: Solution: (3) In a ferromagnetic material, below the curie temperature a domain is defined as a macroscopic region with saturation mag...
Read More →Which term of the AP 20,
Question: Which term of the AP $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$ is its first negative term? Solution: The given $\mathrm{AP}$ is $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots .$ Here, $a=20$ and $d=19 \frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$ Let thenthterm of the given AP be the first negative term. Then, $a_{n}0$ $\Rightarrow 20+(n-1) \times\left(-\frac{3}{4}\right)0 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 20+\frac{3}{4}-\frac{...
Read More →What is the major product formed by HI on reaction with
Question: What is the major product formed by HI on reaction with Correct Option: , 3 Solution:...
Read More →A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:
Question: A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:(1) decrease in size and changes orientation.(2) may increase or decrease in size and change its orientation.(3) increase in size but no change in orientation.(4) have no relation with external magnetic field.Correct Option: , 2 Solution: (2) Atoms of ferromagnetic material in unmagnetised state form domains inside the ferromagnetic material. These domains have large magnetic moment of atoms. In...
Read More →The set of all real values of
Question: The set of all real values of $\lambda$ for which the quadratic equations, $\left(\lambda^{2}+1\right) x^{2}-4 \lambda x+2=0$ always have exactly one root in the interval $(0,1)$ is :(1) $(0,2)$(2) $(2,4]$(3) $(1,3]$(4) $(-3,-1)$Correct Option: , 3 Solution: The given quadratic equation is $\left(\lambda^{2}+1\right) x^{2}-4 \lambda x+2=0$ $\because$ One root is in the interval $(0,1)$ $\therefore f(0) f(1) \leq 0$ $\Rightarrow 2\left(\lambda^{2}+1-4 \lambda+2\right) \leq 0$ $\Rightarr...
Read More →Which term of the AP 121, 117, 113, ... is its first negative term?
Question: Which term of the AP 121, 117, 113, ... is its first negative term? Solution: The given AP is 121, 117, 113, ... .Here,a= 121 andd= 117 121 = 4Let thenthterm of the given AP be the first negative term. Then, $a_{n}0$ $\Rightarrow 121+(n-1) \times(-4)0 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 125-4 n0$ $\Rightarrow-4 n-125$ $\Rightarrow n\frac{125}{4}=31 \frac{1}{4}$ $\therefore n=32$ Hence, the 32nd term is the first negative term of the given AP....
Read More →Which of the following statements are correct?
Question: Which of the following statements are correct? (A) Electric monopoles do not exist whereas magnetic monopoles exist. (B) Magnetic field lines due to a solenoid at its ends and outside cannot be completely straight and confined. (C) Magnetic field lines are completely confined within a toroid. (D) Magnetic field lines inside a bar magnet are not parallel. (E) $\chi=-1$ is the condition for a perfect diamagnetic material, where $\chi$ is its magnetic susceptibility. Choose the correct an...
Read More →Prove the following
Question: If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2 x^{2}+2 q x+1=0$, then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to :(1) $\frac{9}{4}\left(9+q^{2}\right)$(2) $\frac{9}{4}\left(9-q^{2}\right)$(3) $\frac{9}{4}\left(9+p^{2}\right)$(4) $\frac{9}{4}\left(9-p^{2}\right)$Correct Option:...
Read More →Is −150 a term of the AP 11, 8, 5, 2, ...?
Question: Is 150 a term of the AP 11, 8, 5, 2, ...? Solution: The given AP is 11, 8, 5, 2, ... .Here,a= 11 andd= 8 11 = 3Let thenthterm of the given AP be 150. Then, $a_{n}=-150$ $\Rightarrow 11+(n-1) \times(-3)=-150 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow-3 n+14=-150$ $\Rightarrow-3 n=-164$ $\Rightarrow n=\frac{164}{3}=54 \frac{2}{3}$ But, the number of terms cannot be a fraction.Hence, 150 is not a term of the given AP....
Read More →Identify products A and B.
Question: Identify products $\mathrm{A}$ and $\mathrm{B}$. Correct Option: , 2 Solution:...
Read More →Is 184 a term of the AP 3, 7, 11, 15, ...?
Question: Is 184 a term of the AP 3, 7, 11, 15, ...? Solution: The given AP is 3, 7, 11, 15, ... .Here,a= 3 andd= 7 3 = 4Let thenthterm of the given AP be 184. Then, $a_{n}=184$ $\Rightarrow 3+(n-1) \times 4=184 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 4 n-1=184$ $\Rightarrow 4 n=185$ $\Rightarrow n=\frac{185}{4}=46 \frac{1}{4}$ But, the number of terms cannot be a fraction.Hence, 184 is not a term of the given AP....
Read More →A bar magnet of length 14 cm
Question: A bar magnet of length $14 \mathrm{~cm}$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of $18 \mathrm{~cm}$ from the center of the magnet. If $\mathrm{B}_{\mathrm{H}}=0.4 \mathrm{G}$, the magnetic moment of the magnet is $\left(1 \mathrm{G}=10^{-4} \mathrm{~T}\right)$(1) $2.880 \times 10^{3} \mathrm{~J} \mathrm{~T}^{-1}$(2) $2.880 \times 10^{2} \mathrm{~J} \mathrm{~T}^{-1}$(3) $2.880 \mathrm{...
Read More →Find the 6th term from the end of the AP 17, 14, 11, ..., (−40).
Question: Find the 6th term from the end of the AP 17, 14, 11, ..., (40). Solution: Here,a= 17 andd= (14 - 17) = -3,l= (-40) andn= 6 Now,nthterm from the end = [l- (n- 1)d]6thterm from the end = [(-40) - (6 - 1)⨯ (-3)]= [-40 + (5 ⨯ 3)] = (-40 + 15) = -25Hence, the 6th term from the end is -25....
Read More →Let f(x) be a quadratic polynomial such that
Question: Let $f(x)$ be a quadratic polynomial such that $f(-1)+f(2)=0$. If one of the roots of $f(x)=0$ is 3 , then its other root lies in:(1) $(-1,0)$(2) $(1,3)$(3) $(-3,-1)$(4) $(0,1)$Correct Option: 1 Solution: Let $f(x)=a x^{2}+b x+c$ Given: $f(-1)+f(2)=0$ $a-b+c+4 a+2 b+c=0$ $\Rightarrow 5 a+b+2 c=0 \quad \ldots$ (i) and $f(3)=0 \Rightarrow 9 a+3 b+c=0$...(ii) From equations (i) and (ii), $\frac{a}{1-6}=\frac{b}{18-5}=\frac{c}{15-9} \Rightarrow \frac{a}{-5}=\frac{b}{13}=\frac{c}{6}$ Produc...
Read More →Find the 8th term from the end of the AP 7, 10, 13, ..., 184.
Question: Find the 8th term from the end of the AP 7, 10, 13, ..., 184. Solution: Here,a= 7 andd= (10 - 7) = 3,l= 184 andn= 8thfrom the end.Now,nthterm from the end = [l- (n-1)d]8thterm from the end = [184 - (8 - 1) ⨯ 3]= [184 - (7 ⨯ 3)] = (184 - 21) = 163Hence, the 8thterm from the end is 163....
Read More →Find the sum of two middle most terms of the AP
Question: Find the sum of two middle most terms of the AP $-\frac{4}{3},-1, \frac{-2}{3}, \ldots, 4 \frac{1}{3}$. Solution: The given $\mathrm{AP}$ is $-\frac{4}{3},-1, \frac{-2}{3}, \ldots, 4 \frac{1}{3}$. First term, $a=-\frac{4}{3}$ Common difference, $d=-1-\left(-\frac{4}{3}\right)=-1+\frac{4}{3}=\frac{1}{3}$ Suppose there arenterms in the given AP. Then, $a_{n}=4 \frac{1}{3}$ $\Rightarrow-\frac{4}{3}+(n-1) \times\left(\frac{1}{3}\right)=\frac{13}{3} \quad\left[a_{n}=a+(n-1) d\right]$ $\Righ...
Read More →Solve the following
Question: Let $\alpha$ and $\beta$ be the roots of the equation, $5 x^{2}+6 x-2=0$. If $S_{n}=\alpha^{n}+\beta^{n}, n=1,2,3, \ldots$, then :(1) $6 S_{6}+5 S_{5}=2 S_{4}$(2) $6 S_{6}+5 S_{5}+2 S_{4}=0$(3) $5 S_{6}+6 S_{5}=2 S_{4}$(4) $5 S_{6}+6 S_{5}+2 S_{4}=0$Correct Option: , 3 Solution: Since, $\alpha$ and $\beta$ are the roots of the equaton $5 x^{2}+6 x-2=0$ Then, $5 \alpha^{2}+6 \alpha-2=0,5 \beta^{2}+6 \beta-2=0$ $5 \alpha^{2}+6 \alpha=2$ $5 S_{6}+6 S_{5}=5\left(\alpha^{6}+\beta^{6}\right)...
Read More →The total number of
Question: The total number of $\mathrm{C}-\mathrm{C}$ sigma bond $/ \mathrm{s}$ in mesityl oxide $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}\right)$ is______________. (Round off to the Nearest Integer). Solution: (5) Mesityle oxide $\therefore \quad \mathrm{C}_{\sigma} \mathrm{C}=5$...
Read More →Find the middle term of the AP 10, 7, 4, ..., (−62).
Question: Find the middle term of the AP 10, 7, 4, ..., (62). Solution: The given AP is 10, 7, 4, ..., 62.First term,a= 10Common difference,d= 7 10 = 3Suppose there arenterms in the given AP. Then, $a_{n}=-62$ $\Rightarrow 10+(n-1) \times(-3)=-62 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow-3(n-1)=-62-10=-72$ $\Rightarrow n-1=\frac{72}{3}=24$ $\Rightarrow n=24+1=25$ Thus, the given AP contains 25 terms. Middle term of the given AP $=\left(\frac{25+1}{2}\right)$ th term = 13th term= 10 + (13 1...
Read More →Prove the following
Question: Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1$. Let $P_{n}=(\alpha)^{n}+(\beta)^{n}, P_{n-1}=11$ and $P_{n+1}=29$ for some integer $n \geq 1$. Then, the value of $P_{n}^{2}$ is Solution: Given, $\alpha+\beta=1, \alpha \beta=-1$ $\therefore$ Quadratic equation with roots $\alpha, \beta$ is $x^{2}-x-1=0 \Rightarrow \alpha^{2}=\alpha+1$ Multiplying both sides by $\alpha^{n-1} \alpha^{n+1}=\alpha^{n}+\alpha^{n-1}$ (1) Similarly, $\beta^{n+1}=\...
Read More →Given below are two statements:
Question: Given below are two statements: Statement-I : 2 -methylbutane on oxidation with $\mathrm{KMnO}_{4}$ gives 2 methylbutan- $2-o 1$. Statement-II : n-alkanes can be easily oxidised to corresponding alcohol with $\mathrm{KMnO}_{4}$. Choose the correct option :Both statement $\mathrm{I}$ and statement $\Pi$ are correctBoth statement I and statement II are incorrectStatement I is correct but Statement II is incorrectStatement I is incorrect but Statement II is correctCorrect Option: , 3 Solu...
Read More →An electron gun is placed inside a long solenoid of radius R
Question: An electron gun is placed inside a long solenoid of radius $R$ on its axis. The solenoid has $\mathrm{n}$ turns/length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed $v$. If the electron does not hit the surface of the solenoid, maximum possible value of $v$ is (all symbols have their standard meaning): (1) $\frac{e \mu_{0} n \mathrm{IR}}{m}$(2) $\frac{e \mu_{0} n \mathrm{IR}}{2 m}$(3) $\frac{e \mu_{0} n \mathrm{IR}}{4 m}$(4) $\...
Read More →The number of solutions of the equation
Question: The number of solutions of the equation $\log _{4}(x-1)=\log _{2}(x-3)$ is Solution: $\frac{1}{2} \log _{2}(x-1)=\log _{2}(x-3)$ $x-1=(x-3)^{2}$ $x^{2}-6 x+9=x-1$ $x^{2}-7 x+10=0$ $x=2,5$ $X=2 \quad$ Not possible as $\log _{2}(x-3)$ is not defined $\Rightarrow$ No. of solution $=1$...
Read More →Find the middle term of the AP 6, 13, 20, ..., 216.
Question: Find the middle term of the AP 6, 13, 20, ..., 216. Solution: The given AP is 6, 13, 20, ..., 216.First term,a= 6Common difference,d= 13 6 = 7Suppose there arenterms in the given AP. Then, $a_{n}=216$ $\Rightarrow 6+(n-1) \times 7=216 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow 7(n-1)=216-6=210$ $\Rightarrow n-1=\frac{210}{7}=30$ $\Rightarrow n=30+1=31$ Thus, the given AP contains 31 terms. Middle term of the given AP $=\left(\frac{31+1}{2}\right)$ th term = 16th term= 6 + (16 1) 7...
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