Isostructural species are those
Question: Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs. (i) [NF3and BF3] (ii) [BF4-and NH4+] (iii) [BCl3and BrCl3] (iv) [NH3and NO3-] Solution: Option (ii) is the answer....
Read More →Find the equation of all lines of slope zero
Question: Find the equation of all lines of slope zero and that is tangent to the curve $y=\frac{1}{x^{2}-2 x+3}$. Solution: finding the slope of the tangent by differentiating the curve $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(2 \mathrm{x}-2)}{\left(\mathrm{x}^{2}-2 \mathrm{x}+3\right)}$ Now according to question, the slope of all tangents is equal to 0 , so $-\frac{(2 x-2)}{\left(x^{2}-2 x+3\right)}=0$ Therefore the only possible solution is $x=1$ since this point lies on the curve, we can fin...
Read More →Find the equation of all lines
Question: Find the equation of all lines having slope 2 and that are tangent to the curve $y=\frac{1}{x-3}, x \neq 3$. Solution: finding the slope of the tangent by differentiating the curve $\frac{d y}{d x}=-\frac{1}{(x-3)^{2}}$ Now according to question, the slope of all tangents is equal to 2 , so $-\frac{1}{(x-3)^{2}}=2$ $(x-3)^{2}=-\frac{1}{2}$ We can see that LHS is always greater than or equal to 0 , while RHS is always negative. Hence no tangent is possible...
Read More →Define ionisation enthalpy.
Question: Define ionisation enthalpy. Discuss the factors affecting ionisation enthalpy of the elements and its trends in the periodic table. Solution: Ionisation enthalpy is the energy required by an isolated and gaseous atom in its ground state to remove an electron. The effective nuclear charge due to the screening effect, the valence electrons are shielded by the inner core electrons. This effective nuclear charge is less than the actual charge present on the atom. Penetrated orbitals: It is...
Read More →Find the equation
Question: Find the equation of the tangent line to the curve $y=x^{2}-2 x+7$ which is perpendicular to the line $5 y-15 x=13$ Solution: slope of given line is 3 finding the slope of the tangent by differentiating the curve $\frac{d y}{d x}=2 x-2$ $m($ tangent $)=2 x-2$ since both lines are perpendicular to each other $(2 \times-2) \times 3=-1$ $\mathrm{X}=\frac{5}{6}$ since this point lies on the curve, we can find y by substituting $x$ $y=\frac{25}{36}-\frac{10}{6}+7=\frac{217}{36}$ therefore, ...
Read More →Find the equation
Question: Find the equation of the tangent line to the curve $y=x^{2}-2 x+7$ which is parallel to the line $2 x-y+9=0$ Solution: finding the slope of the tangent by differentiating the curve $\frac{d y}{d x}=2 x-2$ $m($ tangent $)=2 x-2$ equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$ now comparing the slope of a tangent with the given equation $\mathrm{m}($ tangent $)=2$ $2 x-2=2$ $x=2$ since this point lies on the curve, we can find $y$ by substituting $...
Read More →Two boats leave a port at the same time.
Question: Two boats leave a port at the same time. One travels 60 km in the direction N 500 E while the other travels 50 km in the direction S 700 E. What is the distance between the boats? Solution: Both the boats starts from $A$ and boat 1 reaches at $B$ and boat 2 reaches at $C$. Here, $A B=60 \mathrm{Km}$ and $A C=50 \mathrm{Km}$ So, the net distance between ta boats is: $|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}-\overrightarrow{\mathrm{AB}}|$ $=\sqrt{60^{2}+50^{2}-2.60 .5...
Read More →Find the equation of a normal to the curve
Question: Find the equation of a normal to the curve $y=x \log _{e} x$ which is parallel to the line $2 x-2 y+3=0$. Solution: finding the slope of the tangent by differentiating the curve $\frac{\mathrm{dy}}{\mathrm{dx}}=\ln \mathrm{x}+1$ $m$ (tangent) $=\ln x+1$ normal is perpendicular to tangent so, $m_{1} m_{2}=-1$] $m($ normal $)=-\frac{1}{\ln x+1}$ equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$ now comparing the slope of normal with the given equation $\mathrm{m}...
Read More →If the angles of a triangle are in the ratio 1 : 2 : 3, prove that its
Question: If the angles of a triangle are in the ratio 1 : 2 : 3, prove that its corresponding sides are in the ratio $1: \sqrt{3}: 2$ Solution: Given: Angles of a triangle are in the ratio 1 : 2 : 3 Need to prove: Its corresponding sides are in the ratio $1: \sqrt{3}: 2$ Let the angles are $x, 2 x, 3 x$ Therefore, $x+2 x+3 x=180^{0}$ $6 x=180^{\circ}$ $x=30^{\circ}$ So, the angles are $30^{\circ}, 60^{\circ}, 90^{\circ}$ So, the ratio of the corresponding sides are: $=\sin 30^{\circ}: \sin 60^{...
Read More →Determine the equation (s) of tangent (s) line to the curve
Question: Determine the equation (s) of tangent (s) line to the curve $y=4 x^{3}-3 x+5$ which are perpendicular to the line $9 y+x+3=0$. Solution: finding the slope of the tangent by differentiating the curve $\frac{d y}{d x}=12 x^{2}-3$ $\mathrm{m}($ tangent $)=12 \mathrm{x}^{2}-3$ the slope of given line is $-\frac{1}{9}$, so the slope of line perpendicular to it is 9 $12 x^{2}-3=9$ $x=1$ or $-1$ since this point lies on the curve, we can find y by substituting $x$ $y=6$ or 4 therefore, the eq...
Read More →Find the equation
Question: Find the equation of normal line to the curve $y=x^{3}+2 x+6$ which is parallel to the line $x+14 y+4=0$. Solution: finding the slope of the tangent by differentiating the curve $\frac{d y}{d x}=3 x^{2}+2$ $\mathrm{m}($ tangent $)=3 x^{2}+2$ normal is perpendicular to tangent so, $m_{1} m_{2}=-1$ $\mathrm{m}($ normal $)=\frac{-1}{3 \mathrm{x}^{2}+2}$ equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$ now comparing the slope of normal with the given equation $\ma...
Read More →In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm,
Question: In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm, find cos A, cos B, cos C. Solution: Given: a = 3 cm, b = 5 cm and c = 7 cm Need to find: cos A, cos B, cos C $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{5^{2}+7^{2}-3^{2}}{2.5 .7}=\frac{65}{70}=\frac{13}{14}$ $\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}=\frac{7^{2}+3^{2}-5^{2}}{2.7 .3}=\frac{33}{42}$ $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{3^{2}+5^{2}-7^{2}}{2.3 .5}=\frac{-15}{30}=-\frac{1}{2}$...
Read More →Solve the triangle in which
Question: Solve the triangle in which $\mathrm{a}=2 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}$ and $\mathrm{c}=\sqrt{3} \mathrm{~cm}$. Solution: Given: $a=2 \mathrm{~cm}, b=1 \mathrm{~cm}$ and $c=\sqrt{3} \mathrm{~cm}$ Perimeter $=a+b+c=3+\sqrt{3} \mathrm{~cm}$ Area $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{\frac{3+\sqrt{3}}{2}\left(\frac{3+\sqrt{3}}{2}-2\right)\left(\frac{3+\sqrt{3}}{2}-1\right)\left(\frac{3+\sqrt{3}}{2}-\sqrt{3}\right)}$ $=\sqrt{\frac{3+\sqrt{3}}{2} \cdot \frac{\sqrt{3}-1}{2} \cdot \f...
Read More →Discuss the factors affecting electron
Question: Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table. Solution: The factors affecting electron gain enthalpy and the trend in its variation in the periodic table are: (1)ATOMIC SIZEAs we go down the group electron gain enthalpy decreases as the distance of the nucleus from the outermost shell increases which decreases its tendency to gain electron and electron gain enthalpy becomes less negative. (2)EFFECTIVE NUCLEAR CHARGEAs we go f...
Read More →Find the equation
Question: Find the equation of the tangent line to the curve $y=x^{2}+4 x-16$ which is parallel to the line $3 x-y+1=0$. Solution: finding the slope of the tangent by differentiating the curve $\frac{d y}{d x}=2 x+4$ $m($ tangent $)=2 x+4$ equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$ now comparing the slope of a tangent with the given equation $2 x+4=3$ $x=-\frac{1}{2}$ Now substituting the value of $x$ in the curve to find $y$ $y=\frac{1}{4}-2-16=-\frac{71}{4}$ T...
Read More →Solve this
Question: In a ∆ABC, if $\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}=\sin ^{2} \mathrm{C}$ show that the triangle is rightangled Solution: Given: $\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}=\sin ^{2} \mathrm{C}$ Need to prove: The triangle is right-angled $\sin ^{2} A+\sin ^{2} B=\sin ^{2} C$ We know, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=2 \mathrm{R}$ So, $\sin ^{2} A+\sin ^{2} B=\sin ^{2} C$ $\frac{a^{2}}{4 R^{2}}+\frac{b^{2}...
Read More →Assertion (A): Electron gain enthalpy becomes less
Question: Assertion (A): Electron gain enthalpy becomes less negative as we go down agroup.Reason (R): Size of the atom increases on going down the group and the addedelectron would be farther from the nucleus. (i) Assertion and reason both are correct statements but the reason is not thecorrect explanation for the assertion. (ii) Assertion and reason both are correct statements and reason is the correct explanationfor the assertion. (iii) Assertion and reason both are wrong statements. (iv) The...
Read More →Assertion (A): Boron has a smaller first ionisation
Question: Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.Reason (R): The penetration of a 2s electron to the nucleus is more than the2p electron hence 2p electron is more shielded by the innercore of electrons than the 2s electrons. (i) Assertion and reason both are correct statements but the reason is not thecorrect explanation for the assertion. (ii) The assertion is a correct statement but the reason is the wrong statement. (iii) Assertion and reason both are corr...
Read More →Electronic configuration of some elements is given
Question: Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpySolution:V. Assertion and Reason TypeIn the following questions, a statement of Assertion (A) followed by a statement ofthe reason (R) is given. Choose the correct option out of the choices given beloweach question.46. Assertion (A): Generally, ionisation enthalpy increases from left to right in a period....
Read More →The equation of the tangent
Question: The equation of the tangent at $(2,3)$ on the curve $y^{2}=a x^{3}+b$ is $y=4 x-5$. Find the values of $a$ and $b$. Solution: finding the slope of the tangent by differentiating the curve $2 y \frac{d y}{d x}=3 a x^{2}$ $\frac{d y}{d x}=\frac{3 a x^{2}}{2 y}$ $\mathrm{m}$ (tangent) at $(2,3)=2 \mathrm{a}$ equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$ now comparing the slope of a tangent with the given equation $2 a=4$ $a=2$ now $(2,3)$ lies on the curve, ...
Read More →Among alkali metals which element
Question: Among alkali metals which element do you expect to be least electronegative and why? Solution: Caesium is the least electronegative alkali metal as electronegativity decreases as we go down the group. Caesium is a group 1 element and lies down the group as it has the largest size due to a decrease in the effective nuclear charge. IV. Matching Type...
Read More →Find the equation
Question: Find the equation of the normal to the curve $a y^{2}=x^{3}$ at the point $\left(a m^{2}, a m^{3}\right)$. Solution: finding the slope of the tangent by differentiating the curve $2 a y \frac{d y}{d x}=3 x^{2}$ $\frac{d y}{d x}=\frac{3 x^{2}}{2 a y}$ $\mathrm{m}$ (tangent) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $\frac{3 \mathrm{~m}}{2}$ normal is perpendicular to tangent so, $m_{1} m_{2}=-1$ $\mathrm{m}$ (normal) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $-\f...
Read More →Solve this
Question: In a ∆ABC, if $\frac{\cos A}{a}=\frac{\cos B}{b}$ show that the triangle is isosceles. Solution: Given: $\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}$ Need to prove: $\triangle \mathrm{ABC}$ is isosceles. $\frac{\cos A}{a}=\frac{\cos B}{b}$ $\Rightarrow \frac{\sqrt{1-\sin ^{2} A}}{a}=\frac{\sqrt{1-\sin ^{2} B}}{b}$ $\Rightarrow \frac{1-\sin ^{2} \mathrm{~A}}{\mathrm{a}^{2}}=\frac{1-\sin ^{2} \mathrm{~B}}{\mathrm{~b}^{2}}$ [Squaring both sides] $\Rightarrow \fra...
Read More →The radius of Na+ cation is less
Question: The radius of Na+ cation is less than that of Na atom. Give reason. Solution: Sodium atom loses one electron to form sodium cation and after the formation of a cation, the effective nuclear charge on the ion increases on the left electrons which results in a decrease in radius....
Read More →Find the equation
Question: Find the equation of the normal to the curve $x^{2}+2 y^{2}-4 x-6 y+8=0$ at the point whose abscissa is 2 Solution: finding slope of the tangent by differentiating the curve $2 x+4 y \frac{d y}{d x}-4-6 \frac{d y}{d x}=0$ $\frac{d y}{d x}=\frac{4-2 x}{4 y-6}$ Finding $y$ co - ordinate by substituting $x$ in the given curve $2 y^{2}-6 y+4=0$ $y^{2}-3 y+2=0$ $y=2$ or $y=1$ $\mathrm{m}$ (tangent) at $\mathrm{x}=2$ is 0 normal is perpendicular to tangent so, $m_{1} m_{2}=-1$ $\mathrm{m}$ (...
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