Find the value of k for which the points A
Question: Find the value of k for which the points A(-2, 3), B(1, 2) and C(k, 0) are collinear. Solution: Given: The points are $A(-5,1), B(1,2)$ and $C(k, 0)$ To find: value of k $A B=\sqrt{(1+5)^{2}+(2-1)^{2}}=\sqrt{36+1}$ $=\sqrt{37}$ units $B C=\sqrt{(k-1)^{2}+4}$ $A C=\sqrt{(k+5)^{2}+1}$ Since the points are collinear, $A B+B C=A C$ $\Rightarrow \sqrt{37}+\sqrt{(\mathrm{k}-1)^{2}+4}=\sqrt{(\mathrm{k}+5)^{2}+1}$ Squaring both sides and rearranging, $\Rightarrow 37+(\mathrm{k}-1)^{2}+4-(\math...
Read More →Consider the reactions given below.
Question: Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (i) to (iv) is correct? (a) C (g) + 4 H (g) CH4 (g); ∆rH = x kJ mol1 (b) C (graphite,s) + 2H2 (g) CH4 (g); ∆rH = y kJ mol1 (i) x = y (ii) x = 2y (iii) x y (iv) x y Solution: Option (iii) x y is the answer....
Read More →On the basis of thermochemical equations (a),
Question: On the basis of thermochemical equations (a), (b) and (c), find out which of the algebraic relationships given in options (i) to (iv) is correct. (a) C (graphite) + O2 (g) CO2 (g) ; ∆rH = x kJ mol1 (b) C (graphite) +12 O2 (g) CO (g) ; ∆rH = y kJ mol1 (c) CO (g) +12 O2 (g) CO2 (g) ; ∆rH = z kJ mol1 (i) z = x + y (ii) x = y z (iii) x = y + z (iv) y = 2z x Solution: Option (iii) x = y + z is the answer....
Read More →The entropy change can be calculated by using
Question: The entropy change can be calculated by using the expression ∆S = qrev/T When water freezes in a glass beaker, choose the correct statement amongst the following : (i) ∆S (system) decreases but ∆S (surroundings) remains the same. (ii) ∆S (system) increases but ∆S (surroundings) decreases. (iii) ∆S (system) decreases but ∆S (surroundings) increases. (iv) ∆S (system) decreases and ∆S (surroundings) also decreases. Solution: Option (iii) is the answer....
Read More →The pressure-volume work for an ideal gas
Question: The pressure-volume work for an ideal gas can be calculated by using the expression w= ʃPexdv. The work can also be calculated from the pV a plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf. choose the correct option. (i) w (reversible) = w (irreversible) (ii) w (reversible) w (irreversible) (iii) w (reversible) w (irreversible) (iv) w (reversible) = w (irreversible) + pex.∆V Solu...
Read More →Show that the points A
Question: Show that the points A(-5, 1), B(5, 5) and C(10, 7) are collinear Solution: Given: The points are A(-5, 1), B(5, 5) and C(10, 7). Note: Three points are collinear if the sum of lengths of any sides is equal to the length of the third side. $\mathrm{AB}=\sqrt{(5+5)^{2}+(5-1)^{2}}=\sqrt{100+16}$ $=2 \sqrt{29}$ units..........(1) $B C=\sqrt{(10-5)^{2}+(7-5)^{2}}=\sqrt{25+4}$ $=\sqrt{29}$ units ...........(2) $A C=\sqrt{(10+5)^{2}+(7-1)^{2}}=\sqrt{225+36}$ $=3 \sqrt{29}$ units ...............
Read More →In an adiabatic process,
Question: In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following. (i) q = 0, ∆T 0, w = 0 (ii) q 0, ∆T = 0, w = 0 (iii) q = 0, ∆T = 0, w = 0 (iv) q = 0, ∆T 0, w 0 Solution: Option (iii)q = 0, ∆T = 0, w = 0 is the answer....
Read More →Show that the following curves intersect orthogonally at the indicated points:
Question: Show that the following curves intersect orthogonally at the indicated points: $x^{2}=4 y$ and $4 y+x^{2}=8$ at $(2,1)$ Solution: Given: Curves $x^{2}=4 y \ldots(1)$ $\ 4 y+x^{2}=8 \ldots(2)$ The point of intersection of two curves $(2,1)$ Solving $(1) \(2)$, we get, First curve is $x^{2}=4 y$c Differentiating above w.r.t $\mathrm{x}$, $\Rightarrow 2 x=4 \cdot \frac{d y}{d x}$ $\Rightarrow \frac{d y}{d x}=\frac{2 x}{4}$ $\Rightarrow m_{1}=\frac{x}{2} \ldots(3)$ Second curve is $4 y+x^{...
Read More →∆fUᶱ of formation of CH4 (g) at certain
Question: ∆fUᶱ of formation of CH4 (g) at certain temperature is 393 kJ mol1. The value of ∆fHᶱ is (i) zero (ii) ∆f Uᶱ (iii) ∆f Uᶱ (iv) equal to ∆f Uᶱ Solution: Option (ii) ∆f Uᶱ is the answer....
Read More →During complete combustion of one mole of butane,
Question: During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is (i) 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l) ∆cH = 2658.0 kJ mol1 (ii) C4H10(g) +13/2 O2 (g) 4CO2 (g) + 5H2O (g) ∆cH = 1329.0 kJ mol1 (iii) C4H10(g) +13/2 O2 (g) 4CO2 (g) + 5H2O (l) ∆cH = 2658.0 kJ mol1 (iv) C4H10 (g) +13/2 O2 (g) 4CO2 (g) + 5H2O (l) ∆cH = +2658.0 kJ mol1 Solution: Option (iii)C4H10(g) +13/2 O2 (g) 4CO2 (g) + 5H2O (l) ∆cH = 2658.0 kJ mo...
Read More →The volume of gas is reduced to half from
Question: The volume of gas is reduced to half from its original volume. The specific heat will be ______. (i) reduce to half (ii) be doubled (iii) remain constant (iv) increase four times Solution: Option (iii) is the answer....
Read More →Find the area of ΔABC whose vertices are A
Question: Find the area of $\triangle \mathrm{ABC}$ whose vertices are $\mathrm{A}(-3,-5), \mathrm{B}(5,2)$ and $\mathrm{C}(-9,-3)$. Solution: Given: The vertices of the triangle are $A(-3,-5), B(5,2)$ and $C(-9,-3)$. Formula: Area of $\triangle A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ Here, $x_{1}=-3, y_{1}=-5$ $x_{2}=5, y_{2}=2$ $x_{3}=-9, y_{3}=-3$ Putting the values, Area of $\triangle \mathrm{ABC}=\frac{1}{2}[-3...
Read More →The state of a gas can be described by quoting
Question: The state of a gas can be described by quoting the relationship between___. (i) pressure, volume, temperature (ii) temperature, amount, pressure (iii) the amount, volume, temperature (iv) pressure, volume, temperature, amount Solution: Option (iv) is the answer....
Read More →Which of the following statements is correct?
Question: Which of the following statements is correct? (i) The presence of reacting species in a covered beaker is an example of an open system. (ii) There is an exchange of energy as well as a matter between the system and the surroundings in a closed system. (iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system. (iv) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system. Solution: Opti...
Read More →If the points A (-2, -1), B(1, 0), C(x, 3) and D(1, y)
Question: If the points $A(-2,-1), B(1,0), C(x, 3)$ and $D(1, y)$ are the vertices of a parallelogram, find the values of x and y. Solution: Given: Vertices of the parallelogram are $A(-2,-1), B(1,0), C(x, 3)$ and $D(1, y)$. To find: values of $x$ and $y$. Since, $A B C D$ is a parallelogram, we have $A B=C D$ and $B C=D A$. $A B=\sqrt{(1+2)^{2}+(0+1)^{2}}=\sqrt{9+1}$ $=\sqrt{10}$ units $B C=\sqrt{(x-1)^{2}+9}$ $C D=\sqrt{(1-x)^{2}+(y-3)^{2}}$ $D A=\sqrt{9+(1+y)^{2}}$ Since $A B=C D$ $\Rightarro...
Read More →Show that the following set of curves intersect orthogonally :
Question: Show that the following set of curves intersect orthogonally : $x^{2}+4 y^{2}=8$ and $x^{2}-2 y^{2}=4$ Solution: Given: Curves $x^{2}+4 y^{2}=8 \ldots(1)$ $\ x^{2}-2 y^{2}=4 \ldots(2)$ Solving (1) \ (2), we get, from 2nd curve, $x^{2}=4+2 y^{2}$ Substituting on $x^{2}+4 y^{2}=8$, $\Rightarrow 4+2 y^{2}+4 y^{2}=8$ $\Rightarrow 6 y^{2}=4$ $\Rightarrow y^{2}=\frac{4}{6}$ $\Rightarrow y=\pm \sqrt{\frac{2}{3}}$ Substituting on $y=\pm \sqrt{\frac{2}{3}}$, we get, $\Rightarrow x^{2}=4+2\left(...
Read More →Thermodynamics is not concerned about______.
Question: Thermodynamics is not concerned about______. (i) energy changes involved in a chemical reaction. (ii) the extent to which a chemical reaction proceeds. (iii) the rate at which a reaction proceeds. (iv) the feasibility of a chemical reaction. Solution: Option (iii) is the answer....
Read More →Explain the term ‘laminar flow’.
Question: Explain the term laminar flow. Is the velocity of molecules the same in all the layers in laminar flow? Explain your answer. Solution: Laminar flow is described when all the fluids (gas or liquid) flow in the layers form. When the liquid flows on the surface the layer of liquid which is in immediate contact with the surface is stationary. The velocity of the subsequent upper layers increases as the distance of the layers increases from the fixed layer which is stationary and in direct ...
Read More →Why does sharp glass edge become
Question: Why does sharp glass edge become smooth on heating it to its melting point in a flame? Explain which property of liquids is responsible for this phenomenon. Solution: Surface tension is the phenomenon responsible for this. The melted glass tends to take the minimum surface area that is sphere or spherical. On heating, the glass melts and it tends to take a round shape at the edge which has a minimum area....
Read More →Why does the boundary between
Question: Why does the boundary between the liquid phase and gaseous phase disappear on heating a liquid to critical temperature in a closed vessel? In this situation what will be the state of the substance? Solution: The boundary between the liquid phase and gaseous phase disappear on heating a liquid up to critical temperature in a closed vessel because at the critical point the densities of liquid and the vapour become equal. The fluid at this stage is called supercritical fluid...
Read More →Show that the following set of curves intersect orthogonally:
Question: Show that the following set of curves intersect orthogonally: $x^{3}-3 x y^{2}=-2$ and $3 x^{2} y-y^{3}=2$ Solution: Given: Curves $x^{3}-3 x y^{2}=-2 \ldots$ (1) $\ 3 x^{2} y-y^{3}=2 \ldots(2)$ Adding (1) \ (2), we get $\Rightarrow x^{3}-3 x y^{2}+3 x^{2} y-y^{3}=-2+2$ $\Rightarrow x^{3}-3 x y^{2}+3 x^{2} y-y^{3}=-0$ $\Rightarrow(x-y)^{3}=0$ $\Rightarrow(x-y)=0$ $\Rightarrow x=y$ Substituting $x=y$ on $x^{3}-3 x y^{2}=-2$ $\Rightarrow x^{3}-3 x x \times x^{2}=-2$ $\Rightarrow x^{3}-3 ...
Read More →Show that the points A
Question: Show that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus. Solution: Given: Vertices of the quadrilateral are $\mathrm{A}(2,-1), \mathrm{B}(3,4), \mathrm{C}(-2,3)$ and $\mathrm{D}(-3,-2)$. Note: For a quadrilateral to be a rhombus, all the sides must be equal in length and the diagonals must not be equal. $A B=\sqrt{(3-2)^{2}+(4+1)^{2}}=\sqrt{1+25}$ $=\sqrt{26 \text { units }}$ $B C=\sqrt{(-2-3)^{2}+(3-4)^{2}}=\sqrt{25+1}$ $=\sqrt{26}$ units $C D=\sqr...
Read More →Assertion (A): Liquids tend to have a maximum
Question: Assertion (A): Liquids tend to have a maximum number of molecules at their surface. Reason (R): Small liquid drops have a spherical shape. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true. Solution: Option (iv) is correct....
Read More →Assertion (A): At the critical temperature,
Question: Assertion (A): At the critical temperature, the liquid passes into a gaseous state imperceptibly and continuously. Reason (R): The density of the liquid and gaseous phase is equal to critical temperature. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true. Solution: Option (i) is correct....
Read More →Assertion (A): Gases do not liquefy above
Question: Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure. Reason (R): Above the critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true. Solution:...
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