Indian Forest Service Examination (IFoS) — Forestry optional · India · Union Public Service Commission
Original practice questions written to the published syllabus · not past papersMeasuring single trees, logs and stands: diameter, height, form, volume, biomass and growth, with the instruments and formulas used in the field.
In the exams: IFoS Forestry Paper II Section A (forest mensuration); ICAR AICE-JRF 59 Unit 4; ICAR AIEEA-PG Group 09 (forest mensuration); PRC Forest Resource Management; SAF Certified Forester; state forester and RFO exams
How our questions for Indian Forest Service Examination spread across the units of Forest Mensuration. Difficulty: 1 easy, 16 medium, 10 hard.
| Growth and increment: CAI, MAI and their relationship | 4 | |
| Angle-count (Bitterlich) sampling and relascopes | 3 | |
| Site index and site quality curves | 3 | |
| Diameter and girth measurement: breast height conventions, calipers, tapes | 2 | |
| Height measurement: principles and hypsometers | 2 | |
| Log volume formulas: Huber, Smalian, Newton | 2 | |
| Volume of standing trees: form factor and form quotient | 2 | |
| Volume tables: local, general and form-class; construction and use | 2 | |
| Tree biomass and allometric equations | 2 | |
| Age determination and stem analysis | 2 | |
| Bark thickness and crown measurements | 1 | |
| Basal area and its uses | 1 | |
| Stand measurement: stems per hectare, basal area, top height, stand volume | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
How is a diameter tape graduated so that diameter can be read directly from it?
B. Each 1 cm of diameter is marked every π cm, about 3.14 cm of tape
Girth = π × diameter, so one centimetre of diameter corresponds to π ≈ 3.14 cm of circumference. Marks every 4 cm would read quarter girth, not diameter.
From a point 25 m (horizontal) from a tree, the angle to the top is 40° above horizontal and to the base 5° below. What is the height?
A. 23.2 m
H = D(tan α + tan β) = 25 × (tan 40° + tan 5°) = 25 × (0.8391 + 0.0875) = 23.2 m. Subtracting the tangents gives 18.8 m and is used only when both readings are on the same side of the horizontal; using sines in place of tangents gives 18.2 m.
Why is basal area per hectare widely used to prescribe thinnings?
C. It joins tree number and size, is quick to measure, and tracks volume and competition
Basal area integrates number and size, can be estimated in seconds with an angle gauge, and is closely related to volume (G × height × form factor) and to competition. It does change after thinning — that is the point of using it.
Using the quarter-girth (Hoppus) rule, what is the volume of a log 4 m long with a mid-girth of 120 cm?
B. 0.360 m³
Hoppus: V = (G/4)² × L = (1.2/4)² × 4 = 0.09 × 4 = 0.360 m³. The true Huber volume from the same girth is G²L/(4π) = 0.458 m³; the Hoppus figure is π/4 of it, a built-in allowance for squaring. Omitting L gives 0.090.
A tree has a stem volume of 0.90 m³, dbh 35 cm and total height 22 m. What is its breast-height form factor?
A. 0.425
f = V / (g × h) = 0.90 / (0.09621 × 22) = 0.90 / 2.1166 = 0.425. Inverting gives 2.35; using π r² with r = 0.35 m gives 0.106.
A local volume table built in one forest division is applied to a plantation of the same species elsewhere where trees of a given diameter are much taller. The likely result is:
A. A biased estimate, since local tables carry their own height-diameter relationship
A one-entry table bakes in the construction stands' heights. Where trees are taller at the same diameter, it will underestimate volume; where shorter, overestimate. A two-entry (general) table or a local table rebuilt from a local height-diameter curve avoids this.
Above-ground biomass of a stand is 100 t/ha and the root-to-shoot ratio is 0.24. What is the total (above- plus below-ground) biomass?
D. 124 t/ha
Below-ground biomass = 100 × 0.24 = 24 t/ha; total = 124 t/ha. Subtracting gives 76; dividing by the ratio gives 416.7.
In stem analysis of a 30-year-old tree, the disc cut at 3.3 m has 24 rings. At approximately what age did the tree reach 3.3 m?
C. 6 years
The rings on a disc record the years since the stem reached that height, so age at that height = tree age − rings = 30 − 24 = 6 years (before any within-section correction). 24 is the number of years since, not the age at which.
Which method detects false and missing rings when ageing trees from ring counts?
C. Cross-dating, matching patterns of wide and narrow rings among trees
Cross-dating aligns distinctive sequences of narrow and wide rings among many trees from the same area, revealing where a ring is extra or absent. It is the basic tool of dendrochronology.
A tree's volume rose from 0.80 m³ to 1.00 m³ in 5 years. What is its volume increment percent by Pressler's formula, p = (200/n)(V2 − V1)/(V2 + V1)?
C. 4.44%
Pressler: p = (200/5) × (0.20/1.80) = 40 × 0.1111 = 4.44%. The compound-interest formula gives 4.56%; simple interest on the initial volume gives 5.00% and on the final volume 4.00%. The stem names Pressler, so only 4.44% is correct.
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