DSSSB PGT Physics Mock Test - 10 - DSSSB TGT/PGT/PRT MCQ

From positions X and Y we conclude that 1, 5, 6 and 3 lie adjacent to 4. Therefore, 2 must lie opposite 4. From positions Y and Z we conclude that 4, 3, 2 and 5 lie adjacent to 6. Therefore, 1 must lie opposite 6. Thus, 2 lies opposite 4, 1 lies opposite 6 and consequently 5 lies opposite 3.

As analysed above, the number on the face opposite 5 is 3. In position X, since 5 lies on the top, therefore 3 must lie at the bottom face.

The figure may be labelled as shown.

Triangles :

The simplest triangles are AEI, EOI, OHI, HAI, EBJ, BFJ, FOJ, OEJ, HOL, OGL, GDL, DHL, OFK, FCK, CGK and GOK i.e. 16 in number.

The triangles composed of two components each are HAE, AEO, EOH, OHA, OEB, EBF, BFO, FOE, DHO, HOG, OGD, GDH, GOF, OFC, FCG and CGO i.e. 16 in number.

The triangles composed of four components each are HEF, EFG, FGH, GHE, ABO, BGO, CDO and DAO i.e. 8 in number.

The triangles composed of eight components each are DAB, ABC, BCD and CDA i.e. 4 in number.

Total number of triangles in the figure = 16 + 16 + 8 + 4 = 44.

Squares :

The squares composed of two components are HIOL, IEJO, JFKO and KGLO i.e. 4 in number.

The squares composed of four components are AEOH, EBFO, OFGC and HOGD i.e.4 in number.

There is only one square EFGH which is composed of eight components.

There is only one square ABCD which is composed of sixteen components.

Total number of squares in the figure = 4 + 4 + 1 + 1 = 10.

The figure may be labelled as shown.

The simplest squares are QUYX, URVY, YVSW and XYWT i.e. 4 in number.

The squares composed of two components each are IMYP, MJNY, YNKO and PYOL i.e. 4 in number.

The squares composed of three components each are AEYH, EBFY, YFCG and HYGD i.e. 4 in number.

There is only one square i.e. QRST composed of four components.

There is only one square i.e. IJKL composed of eight components.

There is only one square i.e. ABCD composed of twelve components.

Total number of squares in the given figure = 4 + 4 + 4+1 + 1 + 1 = 15.

The figure may be labelled as shown.

Triangles:

The simplest triangles are KJN, KJO, CNB, OEF, JIL, JIM, BLA and MFG i.e. 8 in number.

The triangles composed of two components each are CDJ, EDJ, NKO, JLM, JAH and JGH i.e. 6 in number.

The triangles composed of three components each are BKI, FKI, CJA and EJG i.e. 4 in number.

The triangles composed of four components each are CDE and AJG i.e. 2 in number.

The only triangle composed of six components is BKF.

Thus, there are 8 + 6 + 4 + 2 + 1 = 21 triangles in the given figure.

Parallelograms :

The simplest parallelograms are NJLB and JOFM i.e. 2 in number.

The parallelograms composed of two components each are CDKB, DEFK, BIHA and IFGH i.e.4 in number.

The parallelograms composed of three components each are BKJA, KFGJ, CJIB and JEFI i.e.4 in number.

There is only one parallelogram i.e. BFGA composed of four components.

The parallelograms composed of five components each are CDJA, DEGJ, CJHA and JEGH i.e.4 in number.

The only parallelogram composed of six components is CEFB.

The only parallelogram composed of ten components is CEGA.

Thus, there are 2 + 4 + 4 + 1 + 4+ 1 + 1 = 17 parallelograms in the given figure.

(Here note that the squares and rectangles are also counted amongst the parallelograms).

S - 1 = R
T - 2 = R
U - 1 = T
D - 2 = B
E - 1 = D
N - 2 = L
T - 1 = S
STUDENT → RRTBDLS

Similarly,

O - 1 = N
F - 2 = D
F - 1 = E
I - 1 = G
C - 2 = B
E - 1 = C
R - 2 = Q
OFFICER → NDEGBCQ

Since the middle term is not distributed in both the premises, no conclusion can be framed. Secondly, if both the premises begin with "Some" no conclusion can be reached.

The figure may be labelled as shown.

The squares composed of two components each are ABKJ, BCLK, CDEL, LEFG, KLGH and JKHI i.e.6 in number.

There is only one square i.e. CEGK composed of four components.

The squares composed of eight components each are ACGI and BDFH i.e. 2 in number.

There are 6 + 1 + 2 = 9 squares in the figure.

A dimensionless quantity can have unit. for example angle (radian). But oppisite is not true. A unitless quantity can never have dimensions.it is the unit that give dimensions.

Quantities with different dimensions cant be added or subtracted. Also the dimension of the product of the quantities in power is always one. But when two quantities are multiplied the dimension of the product is the product of the dimensions of the initial quantities.

The two physical quantities which do not have the same dimensions can’t be added or subtracted in the same expression because doing so would lead to equating two quantities of different dimensions, which is non possible. Ex [F]  [P].

Correct Answer :- d

Explanation : a) A topological sort of a directed acyclic graph (DAG) is any ordering m1, m2, …, mn of the nodes

of the graph, such that if mimj is an edge then mi appears before mj . Any topological sort of a

the dependency graph gives a valid evaluation order for the semantic rules. 

b) The parse tree can be annotated with synthesized or inherited attributes. The parse tree can also be indicated with an arrow mark to indicate the manner in which the value gets propagated between the nodes of the parse tree. This graph is called as dependency graph as it indicates the dependency between nodes for deriving the values. This graph is an acyclic graph which doesn’t have a cycle. The presence of a cycle indicates that the graph is incorrect as the dependence of nodes for deriving values cannot be predicted. Edges in the dependence graph show the evaluation order for attribute values and thus the graph is a directed one.

We know that E = hv , where E is energy and v is frequency.
Thus we get h = E/v
And [h] = [E/v] = ML²T^-2 / T^-1
= ML²T^-1
[P] = MLT^-1
[F] = MLT^-2
[Angular momentum] = [P x r] = ML²T^-1

As the power of 10 is irrelevant in the determination of significant figures hence, the numbers of significant figures are 5 i.e. 9,4,6,0,5

As the power of 10 is irrelevant to the determination of significant figures hence, the numbers of significant figures are 3 (1, 8 and 4).

The multiplication of the numbers:
10.610 × 0.210 = 2.2281

Since, the numbers 10.610 and 0.210 have 5 and 3 significant digits, respectively. So, the final result must also have 3 significant digits.

The final answer is 2.23 after rounding off 2.2281. 

• As per rule, zero between two non zero digits are significant.
• The number of a significant figure in the length of the side is 4.
• So the calculated volume should be rounded off to 4 significant figures. Multiplying numbers do not increase significant figures.

Therefore, the correct answer is 4. 

Step 1: Given that:
A particle goes from point A to B on the semicircular path.
Radius of the path(r) = 1.0m
Time taken(t) = 1sec

Step 2: calculation of the average velocity of the particle:
Average velocity = Total displacement/Total time taken
Total displacement = Shortest distance between initial and final point
= Diameter of the semicircle
= 2 × Radius of semicircle
= 2 × 1m
= 2m
Thus,
Average velocity = 2m/1sec
= 2ms⁻¹
Thus,
Option B) 2ms⁻¹ is the correct option.

Since the object does not change its direction it means that the object is traveling in a straight line.
⇒ The path length and displacement will be equal.

1 km = 1000 m and 1 hr = 3600 sec.

Therefore,

From the figure,

dy/dt = u
tanθ = h/x = H/(y+x) ⇒ x = hy/(H-h)
So, dx/dt = h/H - h dy/dt = hu/(H-h)
Now the distance of the edge of shadow from the stationary lamp-post is x + y
So, the speed of the edge of the shadow on ground would be,
dx/dt + dy/dt = u + hu/(H − h) = Hu/(H − h)

In the given all options we can see that curves in graph A and D meet in the first quadrant while the curve in option b meets in the second quadrant but those of option c are parallel and hence they never meet.