The area of a triangle is 5. Two of its vertices are (2, 1)
Question: The area of a triangle is 5. Two of its vertices are (2, 1) and (3, 2). The third vertex lies on y = x + 3. Find the third vertex. Solution: GIVEN: The area of triangle is 5.Two of its vertices are (2, 1) and (3, 2). The third vertex lies ony=x+3 TO FIND: The third vertex. PROOF: Let the third vertex be (x,y) We know area of triangle formed by three points $\left(x_{1,} y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|\left(x_...
Read More →If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k.
Question: If the pointP(2, 2) is equidistant from the pointsA(2,k) andB(2k, 3), findk. Also find the length ofAP. Solution: As per the question, we have $A P=B P$ $\Rightarrow \sqrt{(2+2)^{2}+(2-k)^{2}}=\sqrt{(2+2 k)^{2}+(2+3)^{2}}$ $\Rightarrow \sqrt{(4)^{2}+(2-k)^{2}}=\sqrt{(2+2 k)^{2}+(5)^{2}}$ $\Rightarrow 16+4+k^{2}-4 k=4+4 k^{2}+8 k+25 \quad$ (Squaring both sides) $\Rightarrow k^{2}+4 k+3=0$ $\Rightarrow(k+1)(k+3)=0$ $\Rightarrow k=-3,-1$ Now for $k=-1$ $A P=\sqrt{(2+2)^{2}+(2-k)^{2}}$ $=\...
Read More →The standard heat of formation
Question: The standard heat of formation $\left(\Delta_{f} \mathrm{H}_{298}^{0}\right)$ of ethane (in $\mathrm{kJ} / \mathrm{mol}$ ), if the heat of combustion of ethane, hydrogen and graphite are $-1560,-393.5$ and $-286 \mathrm{~kJ} / \mathrm{mol}$, respectively is __________________ . Solution: $(-192.5)$ $\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} ; \Delta_{\mathrm{C}} \mathrm{H}^{0}=-286.0 \mathrm{~kJ} / \mathrm{mol}$ ...(i) $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow ...
Read More →The point A divides the join of P (−5, 1) and Q(3, 5) in the ratio k:1.
Question: The point A divides the join ofP(5, 1) andQ(3, 5) in the ratiok:1. Find the two values ofkfor which the area of ΔABCwhereBis (1, 5) andC(7, 2) is equal to 2 units. Solution: GIVEN: point A divides the line segment joining P (5, 1) and Q (3, 5) in the ratiok: 1 Coordinates of point B (1, 5) and C (7, 2) TO FIND: The value ofk PROOF: point A divides the line segment joining P (5, 1) and Q (3, 5) in the ratiok: 1 So the coordinates of $A$ are $\left[\frac{3 k+1}{k+1}, \frac{5 k+1}{k+1}\ri...
Read More →For the reaction;
Question: For the reaction; $\mathrm{A}(\mathrm{l}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})$ $\Delta U=2.1 \mathrm{kcal}, \Delta S=20 \mathrm{cal} K^{-1}$ at $300 \mathrm{~K}$ Hence $\Delta \mathrm{G}$ in $\mathrm{kcal}$ is ___________________ . Solution: $(-2.70)$ $\Delta \mathrm{U}=2.1 \mathrm{kcal}=2.1 \times 10^{3} \mathrm{cal}$ $\Delta \mathrm{n}_{\mathrm{g}}=2$ $\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta n_{\mathrm{g}} \mathrm{RT}$ $=2.1 \times 10^{3}+2 \times 2 \times 300$ $=2100+1200$ $=3...
Read More →If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3),
Question: If the pointP(2, 2) is equidistant from the pointsA(2,k) andB(2k, 3), findk. Also find the length ofAP. Solution: As per the question, we have $A P=B P$ $\Rightarrow \sqrt{(2+2)^{2}+(2-k)^{2}}=\sqrt{(2+2 k)^{2}+(2+3)^{2}}$ $\Rightarrow \sqrt{(4)^{2}+(2-k)^{2}}=\sqrt{(2+2 k)^{2}+(5)^{2}}$ $\Rightarrow 16+4+k^{2}-4 k=4+4 k^{2}+8 k+25 \quad$ (Squaring both sides) $\Rightarrow k^{2}+4 k+3=0$ $\Rightarrow(k+1)(k+3)=0$ $\Rightarrow k=-3,-1$ Now for $k=-1$ $A P=\sqrt{(2+2)^{2}+(2-k)^{2}}$ $=\...
Read More →Prove that the points (a, 0), (0, b) and
Question: Prove that the points $(a, 0),(0, b)$ and $(1,1)$ are collinear if $\frac{1}{a}+\frac{1}{b}=1$. Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3} y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_...
Read More →For a reaction,
Question: For a reaction, $4 \mathrm{M}(\mathrm{s})+\mathrm{nO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{M}_{2} \mathrm{O}_{\mathrm{n}}(\mathrm{s})$, the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :The slope change from negative to positiveThe free energy change shows a change from negative to positive valueThe slop changes from positive to negativeThe slop changes from positive t...
Read More →Show that the following sets of points are collinear.
Question: Show that the following sets of points are collinear. (a) $(2,5),(4,6)$ and $(8,8)$ (b) $(1,-1),(2,1)$ and $(4,5)$ Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3} y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\r...
Read More →The variation of equilibrium constant with temperature is given below:
Question: The variation of equilibrium constant with temperature is given below: The values of $\Delta \mathrm{H}^{\circ}, \Delta \mathrm{G}^{\circ}$ at $\mathrm{T}_{1}$ and $\Delta \mathrm{G}^{\circ}$ at $\mathrm{T}_{2}$ (in $\mathrm{kj} \mathrm{mol}^{-1}$ ) respectively, are close to [use $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]$28.4,-7.14$ and $-5.71$$0.64,-7.14$ and $-5.71$$28.4,-5.71$ and $-14.29$$0.64,-5.71$ and $-14.29$Correct Option: , 3 Solution: $\Delta \mat...
Read More →If the point C(− 2, 3) is equidistant from the points A(3, − 1) and B(x, 8),
Question: If the pointC( 2, 3) is equidistant from the pointsA(3, 1) andB(x, 8), find the value ofx. Also, find the distanceBC. Solution: As per the question, we have $A C=B C$ $\Rightarrow \sqrt{(-2-3)^{2}+(3+1)^{2}}=\sqrt{(-2-x)^{2}+(3-8)^{2}}$ $\Rightarrow \sqrt{(5)^{2}+(4)^{2}}=\sqrt{(x+2)^{2}+(-5)^{2}}$ $\Rightarrow 25+16=(x+2)^{2}+25 \quad$ (Squaring both sides) $\Rightarrow 25+16=(x+2)^{2}+25$ $\Rightarrow(x+2)^{2}=16$ $\Rightarrow x+2=\pm 4$ $\Rightarrow x=-2 \pm 4=-2-4,-2+4=-6,2$ Now $B...
Read More →The vertices of ΔABC are (−2, 1), (5, 4) and (2, −3) respectively.
Question: The vertices of ΔABCare (2, 1), (5, 4) and (2, 3) respectively. Find the area of the triangle and the length of the altitude throughA. Solution: GIVEN: The vertices of triangle ABC are A (2, 1) and B (5, 4) and C (2, 3) TO FIND: The area of triangle ABC and length if the altitude through A PROOF: We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3...
Read More →For a dimerization reaction,
Question: For a dimerization reaction, $2 \mathrm{~A}(\mathrm{~g}) \rightarrow \mathrm{A}_{2}(\mathrm{~g})$, at $298 \mathrm{~K}, \Delta \mathrm{U}^{\Theta}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}^{\Theta}=-30 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, then the $\Delta \mathrm{G}^{\Theta}$ will be ________________J. Solution: $(-13538)$ From $\Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ $\Delta \mathrm{H}^{\circ}=-20 \times...
Read More →If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), find the value of x.
Question: If the pointsA(4, 3) andB(x, 5) lie on a circle with centreO(2, 3), find the value ofx. Solution: Given, the pointsA(4, 3) andB(x, 5) lie on a circle with centreO(2, 3).ThenOA=OBAlso (OA)2= (OB)2 $\Rightarrow(4-2)^{2}+(3-3)^{2}=(x-2)^{2}+(5-3)^{2}$ $\Rightarrow(2)^{2}+(0)^{2}=(x-2)^{2}+(2)^{2}$ $\Rightarrow 4=(x-2)^{2}+4$ $\Rightarrow(x-2)^{2}=0$ $\Rightarrow x-2=0$ $\Rightarrow x=2$ Therefore,x= 2....
Read More →The four vertices of a quadrilateral are (1, 2), (−5, 6), (7, −4) and (k, −2) taken in order.
Question: The four vertices of a quadrilateral are (1, 2), (5, 6), (7, 4) and (k, 2) taken in order. If the area of the quadrilateral is zero, find the value ofk. Solution: GIVEN: The four vertices of quadrilateral are (1, 2), (5, 6), (7, 4) and D (k, 2) taken in order. If the area of the quadrilateral is zero TO FIND: value ofk PROOF: Let four vertices of quadrilateral are A (1, 2) and B (5, 6) and C (7, 4) and D (k, 2) We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),...
Read More →Five moles of an ideal gas at 1 bar and 298 k is expanded into vacuum to double the volume.
Question: Five moles of an ideal gas at 1 bar and $298 \mathrm{~K}$ is expanded into vacuum to double the volume. The work done is :$\mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$$-\mathrm{RT}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$$-R T \ln V_{1} / V_{1}$zeroCorrect Option: , 4 Solution: In expansion against vacuum $P_{\text {ext }}=0$ $W=-P_{\text {ext }} \Delta V=0$ w = 0...
Read More →Find the coordinates of the point equidistant from the three points A(5, 3), B(5, −5) and C(1, −5).
Question: Find the coordinates of the point equidistant from the three pointsA(5, 3),B(5, 5) andC(1, 5). Solution: Let the required point beP(x,y). ThenAP=BP=CPThat is, (AP)2=(BP)2=(CP)2This means (AP)2=(BP)2 $\Rightarrow(x-5)^{2}+(y-3)^{2}=(x-5)^{2}+(y+5)^{2}$ $\Rightarrow x^{2}-10 x+25+y^{2}-6 y+9=x^{2}-10 x+25+y^{2}+10 y+25$ $\Rightarrow x^{2}-10 x+y^{2}-6 y+34=x^{2}-10 x+y^{2}+10 y+50$ $\Rightarrow x^{2}-10 x+y^{2}-6 y-x^{2}+10 x-y^{2}-10 y=50-34$ $\Rightarrow-16 y=16$ $\Rightarrow y=-\frac{...
Read More →Find the area of the quadrilaterals, the coordinates of whose vertices are
Question: Find the area of the quadrilaterals, the coordinates of whose vertices are (i) $(-3,2),(5,4),(7,-6)$ and $(-5,-4)$ (ii) $(1,2),(6,2),(5,3)$ and $(3,4)$ (iii) $(-4,-2,(-3,-5),(3,-2),(2,3)$ Solution: (i) Let the vertices of the quadrilateral be A (3, 2), B (5, 4), C (7, 6), and D (5, 4). Join AC to form two trianglesΔABC andΔACD. Area of a triangle $=\frac{1}{2}\left\{x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right\}$ Area of $\triangle \ma...
Read More →For one mole of an ideal gas,
Question: For one mole of an ideal gas, which of these statements must be true? (a) $\mathrm{U}$ and $\mathrm{H}$ each depends only on temperature (b) Compressibility factor $z$ is not equal to 1 (c) $\mathrm{C}_{\mathrm{P}, \mathrm{m}}-\mathrm{C}_{\mathrm{V}, \mathrm{m}}=\mathrm{R}$ (d) $\mathrm{dU}=\mathrm{C}_{\mathrm{V}} \mathrm{dT}$ for any process(1) (a) and (c)(b), (c) and (d)(c) and (d)(a), (c) and (d)Correct Option: , 4 Solution: (a) For ideal gas $\mathrm{U}$ and $\mathrm{H}$ are functi...
Read More →The heat of combustion of ethanol into carbon dioxids and water is
Question: The heat of combustion of ethanol into carbon dioxids and water is $-327 \mathrm{kcal}$ at constant pressure. The heat evolved (in cal) at constant volume and $27^{\circ} \mathrm{C}$ (if all gases behave ideally) is $\left(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$ ______ Solution: $(-326400)$ $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(l)$ $\Delta \mathrm{H...
Read More →The internal energy change
Question: The internal energy change (in J) when $90 \mathrm{~g}$ of water undergoes complete evaporation at $100^{\circ} \mathrm{C}$ is_____________ (Given: $\Delta \mathrm{H}_{\text {vap }}$ for water at $373 \mathrm{~K}=41 \mathrm{~kJ} / \mathrm{mol}, \mathrm{R}=8.314 \mathrm{JK}^{-}$ ${ }^{1} \mathrm{~mol}^{-1}$ ) Solution: $(189494)$ $\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta n_{\mathrm{g}} \mathrm{RT}$ $n=\frac{90}{18}=5 \mathrm{~mol}$ $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharp...
Read More →Find the area of a triangle whose vertices are
Question: Find the area of a triangle whose vertices are (i) $(6,3)(-3,5)$ and $(4,-2)$ (ii) $\left(a t_{1}^{2}, 2 a t_{1}\right),\left(a t_{2}^{2}, 2 a t_{2}\right)$ and $\left(a t_{3}^{2}, 2 a t_{3}\right)$ (iii) $(a, c+a),(a, c)$ and $(-a, c-a)$ Solution: We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left...
Read More →If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that
Question: IfP(x,y) is a point equidistant from the pointsA(6, 1) andB(2, 3), show thatxy= 3. Solution: The given points areA(6, 1) andB(2, 3). The pointP(x,y) is equidistant from the pointsAandB.So,PA=PB.Also, (PA)2= (PB)2 $\Rightarrow(6-x)^{2}+(-1-y)^{2}=(2-x)^{2}+(3-y)^{2}$ $\Rightarrow x^{2}-12 x+36+y^{2}+2 y+1=x^{2}-4 x+4+y^{2}-6 y+9$ $\Rightarrow x^{2}+y^{2}-12 x+2 y+37=x^{2}+y^{2}-4 x-6 y+13$ $\Rightarrow x^{2}+y^{2}-12 x+2 y-x^{2}-y^{2}+4 x+6 y=13-37$ $\Rightarrow-8 x+8 y=-24$ $\Rightarro...
Read More →For a chemical reaction
Question: For a chemical reaction $A+B \rightleftharpoons C+D\left(\Delta_{r} \mathrm{H}^{\ominus}=80 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ the entropy change $\Delta_{r} \mathrm{~S}^{\ominus}$ depends on the temperature $\mathrm{T}$ (in $\mathrm{K}$ ) as $\Delta_{r} S^{\ominus}=2 T\left(J \mathrm{~K}^{-1}\right.$ mol $\left.^{-1}\right)$ Minimum temperature at which it will become spontaneous is ________________$\mathrm{K}$. Solution: (200) $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}...
Read More →If the point P(x, y) is equidistant from the points A(5, 1)
Question: If the pointP(x,y) is equidistant from the pointsA(5, 1) andB( 1, 5), prove that 3x= 2y. Solution: As per the question, we have $A P=B P$ $\Rightarrow \sqrt{(x-5)^{2}+(y-1)^{2}}=\sqrt{(x+1)^{2}+(y-5)^{2}}$ $\Rightarrow(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2} \quad$ (Squaring both sides) $\Rightarrow x^{2}-10 x+25+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-10 y+25$ $\Rightarrow-10 x-2 y=2 x-10 y$ $\Rightarrow 8 y=12 x$ $\Rightarrow 3 x=2 y$ Hence, 3x =2y....
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